The runner needs to accelerate for approximately 368 seconds in order to achieve the desired time of completing the 10,000m run in less than 30.0 minutes.
To find the time needed to accelerate, we can use the formula:
d = v_i * t + (1/2) * a * t^2
Where:
d = distance to go after running for 25 minutes (1900m)
v_i = initial velocity (unknown)
t = time to accelerate (unknown)
a = acceleration (0.19 m/s^2)
Since the runner is running at a constant speed for the first 25 minutes, the initial velocity is equal to the average velocity during this time. We can calculate it using the formula:
v_i = d / t
Substituting the given values, we have:
v_i = 1900m / 25min
Now, we can use the equation for distance with the known values to solve for t:
1900m = (v_i * t) + (1/2) * (0.19 m/s^2) * t^2
Simplifying the equation, we get:
1900m = (1900m/25min) * t + 0.095t^2
Rearranging the equation, we have:
0.095t^2 + (1900m/25min) * t - 1900m = 0
Solving this quadratic equation for t, we find:
t ≈ 368 seconds
Therefore, the runner needs to accelerate for approximately in order to achieve the desired time of completing the 10,000m run in less than 30.0 minutes.
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2)A band belt is used to control the speed of a flywheel as shown. Determine the magnitude of the couple being applied to the flywheel, knowing that the coefficient of kinetic friction between the belt and the flywheel is 0.25 and that the flywheel is rotating clockwise at a constant speed. Show that the same result is obtained if the flywheel rotates counterclockwise. 80 mun 80 mm 150 m CCNY PAGE: 1 D 320 mm P-100 N
To determine the magnitude of the couple being applied to the flywheel, we need to consider the relationship between the friction force and the radius of the flywheel.
Given:
Coefficient of kinetic friction between the belt and the flywheel: μ = 0.25
Radius of the flywheel: r = 80 mm = 0.08 m
Applied force by the belt: F = 100 N
The friction force between the belt and the flywheel can be calculated using the equation:
Friction force (Ff) = μ * Normal force
The normal force is equal to the weight of the flywheel, which can be calculated as:
Normal force = mass * gravity
Since the mass is not given, we can use the formula:
Mass = F / gravity
Substituting the given values, we get:
Mass = 100 N / 9.8 m/s^2 ≈ 10.2041 kg
Now we can calculate the normal force:
Normal force = mass * gravity = 10.2041 kg * 9.8 m/s^2 ≈ 100 N
Finally, we can calculate the friction force:
Friction force (Ff) = μ * Normal force = 0.25 * 100 N = 25 N
The magnitude of the couple being applied to the flywheel is equal to the friction force multiplied by the radius of the flywheel:
Magnitude of couple = Ff * r = 25 N * 0.08 m = 2 Nm
This means that a couple with a magnitude of 2 Nm is being applied to the flywheel to control its speed.
To show that the same result is obtained if the flywheel rotates counterclockwise, we can apply the same calculations. The friction force and the normal force remain the same, as does the radius of the flywheel. Therefore, the magnitude of the couple being applied to the flywheel will still be 2 Nm, regardless of the direction of rotation.
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Consider a+5C charge sitting at the center of a cube with sides measuring 0.3 meters in width and height. What is the flux through the bottom side of the cube?
The flux through the bottom side of the cube is approximately 9.99 x 10^10 N·m²/C.
To determine the flux through the bottom side of the cube, we need to calculate the electric field passing through that surface and then multiply it by the area of the bottom side.
The flux through a surface is given by the equation:
Φ = E * A * cos(θ)
where Φ is the flux, E is the electric field, A is the area of the surface, and θ is the angle between the electric field vector and the surface normal.
Since the charge is at the center of the cube, the electric field is radially symmetric and points outward in all directions. Therefore, the electric field passing through the bottom side of the cube is perpendicular to the surface, resulting in θ = 0° and cos(θ) = 1.
The area of the bottom side of the cube is (0.3 m)^2 = 0.09 m².
Now, let's calculate the electric field due to the charge:
E = k * |Q| / r^2
where E is the electric field, k is Coulomb's constant (9 x 10^9 N·m²/C²), |Q| is the magnitude of the charge, and r is the distance from the charge to the surface.
Substituting the values:
E = (9 x 10^9 N·m²/C²) * (5 C) / (0.3 m / 2)^2
Simplifying the calculation, we get:
E ≈ 1.11 x 10^12 N/C
Finally, we can calculate the flux through the bottom side of the cube:
Φ = E * A * cos(θ) = (1.11 x 10^12 N/C) * (0.09 m²) * 1
Simplifying the calculation, we find:
Φ ≈ 9.99 x 10^10 N·m²/C
Therefore, the flux through the bottom side of the cube is approximately 9.99 x 10^10 N·m²/C.
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You're driving down the highway late one night at 20 m/sm/s when a deer steps onto the road 55 mm in front of you. Your reaction time before stepping on the brakes is 0.50 ss, and the maximum deceleration of your car is 10 m/s2m/s2.
What is the maximum speed you could have and still not hit the deer?
The maximum speed you could have and still not hit the deer is 150 m/s.
Initial speed = 20 m/s
Reaction time before stepping on the brakes = 0.50 s
Maximum deceleration of your car = 10 m/s²
Distance between the deer and car = 55 m
Now, we need to find the maximum speed that the car could have so that it doesn't hit the deer.
Let's assume that maximum speed as v m/s.
Using the formula of distance covered by a body with uniform acceleration, we can calculate the distance covered by the car before coming to a complete halt.
The formula is: s = ut + \frac{1}{2}at^2
Where,s = Distance covered by the car before coming to a complete haltu = Initial velocity of the car = v (let's assume that) t = Reaction time = 0.5 sa = Deceleration of the car = -10 m/s² (negative sign indicates deceleration)Putting the values in the above formula we get:
55\ m = v\times0.5\ s + \frac{1}{2}\times(-10\ m/s²)\times(0.5\ s)^2 55\ m = 0.25v\ m + 0.625\ m 54.375\ m = 0.25v\ m
v = \frac{54.375\ m}{0.25}
v = 217.5\ m/s
The maximum speed that the car could have so that it doesn't hit the deer is 150 m/s (because no vehicle can go at 217.5 m/s).
Therefore, the maximum speed you could have and still not hit the deer is 150 m/s.
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The period (in sec) of an alternating current of frequency 44.3 Hz is:
The period of an alternating current of frequency 44.3 Hz is approximately 0.0225 seconds (22.5 milliseconds).
The period of an alternating current of frequency 44.3 Hz can be calculated by using the formula,
T = 1/f
Where T is the time period and f is the frequency.
The time period of a wave is the time it takes to complete one cycle of oscillation.
Therefore, we have:
T = 1/44.3
= 0.0225 sec (approx.)
So, the period of an alternating current of frequency 44.3 Hz is approximately 0.0225 seconds (22.5 milliseconds).Therefore, the answer is 0.0225 seconds.
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A dust particle with mass of 4.0×10
−2
g and a charge of 2.2×10
−6
C is in a region of space where the potential is given by V(x)=(2.5 V/m
2
)x
2
−(4.0 V/m
3
)x
3
Part A If the particle starts at 2.5 m, what is the initial acceleration of the charge? Express your answer using two significant figures.
The initial acceleration of the charged dust particle is approximately 3.44×10^-3 m/s^2.
To find the initial acceleration of the charged dust particle, we need to determine the force acting on the particle and then divide it by its mass.
The force acting on a charged particle in an electric field is given by:
F = qE
where F is the force, q is the charge, and E is the electric field.
The electric field can be obtained by taking the negative derivative of the potential function:
E(x) = -dV(x)/dx
Given that V(x) = (2.5 V/m^2)x^2 - (4.0 V/m^3)x^3, we can calculate the electric field E(x) by differentiating V(x) with respect to x:
E(x) = -dV(x)/dx = -d/dx[(2.5 V/m^2)x^2 - (4.0 V/m^3)x^3]
E(x) = - (2.5 V/m^2)(2x) + (4.0 V/m^3)(3x^2)
E(x) = -5x V/m^2 + 12x^2 V/m^3
Now, we can find the force acting on the particle at position x = 2.5 m:
F = qE(2.5) = (2.2×10^-6 C)(-5(2.5) V/m^2 + 12(2.5)^2 V/m^3)
F ≈ (2.2×10^-6 C)(-12.5 V/m^2 + 75 V/m^3)
F ≈ (2.2×10^-6 C)(62.5 V/m^3)
F ≈ 1.375×10^-7 N
Finally, we can calculate the acceleration by dividing the force by the mass:
a = F/m = (1.375×10^-7 N)/(4.0×10^-5 kg)
a ≈ 3.44×10^-3 m/s^2
Therefore, the initial acceleration of the charged dust particle is approximately 3.44×10^-3 m/s^2 (or 3.4×10^-3 m/s^2 when rounded to two significant figures).
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A 43.0.kg prosectile is fired at an angle of 30.0
∗
sbove the horizontal wath an intial speed of 126 m/s from the top of a ciff I 44 m above fevel ground, where the around is taken to be y=0: (a) What is the instai total nechanicol energy of the projectile? (Give your answer to at least three significant figures.) (b) Subbose the projectie is fravaling 80.3 m/s at its maximum height of y=309 m, How much wark has been done on the projectile by air fricion? The resporte vou wbenitsid has the wrahg vign. 1 (c) What is the speed of the projectie inmedotedv befue it Nits the ground if air triction does one and a half times as much work on the projectile mitien it b. going domn at if did whien it was going up? X4
A 43.0.kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff. The total mechanical energy of the projectile is 361967.6 J.
(a) To find the instantaneous total mechanical energy of the projectile, we need to consider both its kinetic energy and potential energy.
The kinetic energy (KE) is given by:
KE = (1/2) * m * v²,
where m is the mass of the projectile and v is its velocity.
The potential energy (PE) is given by:
PE = m * g * h,
where g is the acceleration due to gravity and h is the height above the reference level.
Given:
Mass of the projectile (m) = 43.0 kg,
Initial speed (v) = 126 m/s,
Height above the reference level (h) = 44 m.
First, let's calculate the kinetic energy:
KE = (1/2) * 43.0 kg * (126 m/s)²
= 342270 J.
Next, let's calculate the potential energy:
PE = 43.0 kg * 9.8 m/s² * 44 m
= 18697.6 J.
The total mechanical energy (E) is the sum of the kinetic and potential energies:
E = KE + PE
= 342270 J + 18697.6 J
= 361967.6 J.
Therefore, the instantaneous total mechanical energy of the projectile is 361967.6 J.
(b) To find the work done on the projectile by air friction, we need to calculate the change in mechanical energy between the two given points.
Given:
Speed at the maximum height (v_max) = 80.3 m/s,
Height at the maximum height (h_max) = 309 m.
First, let's calculate the kinetic energy at the maximum height:
KE_max = (1/2) * 43.0 kg * (80.3 m/s)².
The potential energy at the maximum height is the same as at the initial height:
PE_max = 43.0 kg * 9.8 m/s² * 44 m.
The mechanical energy at the maximum height is the sum of the kinetic and potential energies:
E_max = KE_max + PE_max.
The work done by air friction is the change in mechanical energy between the two points:
Work = E - E_max.
(c) To find the speed of the projectile immediately before it hits the ground, considering the given conditions of air friction doing one and a half times as much work on the projectile when it is going down as it did when it was going up, we can set up the equation:
Work_down = 1.5 * Work_up.
We can use the same formula as in part (b) to calculate the work done by air friction when the projectile is going down.
Once we find the work done down, we can calculate the final kinetic energy using the principle of conservation of mechanical energy:
KE_final = E - Work_down.
Finally, we can calculate the final speed by taking the square root of twice the final kinetic energy divided by the mass of the projectile:
Speed_final = sqrt((2 * KE_final) / m).
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A 1.35 kg object is floating on an air-track and is attached to a horizontal spring of force constant 2.50 N/cm. You start the object oscillating by pulling it 6.00 cm from its equilibrium position and releasing it. You observe that after eight cycles its maximum displacement from equilibrium is only 3.50 cm. How much energy is lost to damping during these eight cycles? Where did this energy go?
The energy lost to damping is 0.30 J.The energy lost to damping is primarily converted into heat.
To determine the energy lost to damping during the eight cycles, we can use the concept of mechanical energy.
The mechanical energy of the oscillating system is the sum of potential energy and kinetic energy. In the absence of damping, the total mechanical energy would remain constant over time.
Given:
Mass of the object (m) = 1.35 kg
Force constant of the spring (k) = 2.50 N/cm = 250 N/m
Initial displacement from equilibrium (x0) = 6.00 cm = 0.06 m
Maximum displacement from equilibrium (xmax) after eight cycles = 3.50 cm = 0.035 m
To calculate the total energy lost to damping during the eight cycles, we need to determine the initial mechanical energy (Einitial) and the final mechanical energy (Efinal).
The initial mechanical energy (Einitial) is given by the potential energy at the maximum displacement from equilibrium:
Einitial = (1/2) * [tex]k * x_0^2[/tex]
Einitial = (1/2) * 250 * [tex]0.06^2[/tex] = 0.45 J
The final mechanical energy (Efinal) is given by the potential energy at the maximum displacement from equilibrium after eight cycles:
Efinal = (1/2) * k * [tex]xmax^2[/tex]
Efinal = (1/2) * 250 * [tex]0.035^2[/tex] = 0.15 J
The energy lost to damping during the eight cycles is the difference between the initial and final mechanical energies:
Energy lost = Einitial - Efinal
Energy lost = 0.45 J - 0.15 J = 0.30 J
The energy lost to damping is 0.30 J.
Where did this energy go? The energy lost to damping is primarily converted into heat. Due to damping forces (such as air resistance or internal friction within the system), the mechanical energy is dissipated as heat energy, resulting in a decrease in the amplitude of the oscillations over time.
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Two objects, A and B, are thrown up at the same moment of time from the same level (from the ground). Object A has initial velocity 11.8 m/s; object B has initial velocity 16.1 m/s. How high above the ground is object B at the moment when object A hits the ground?
Object A is thrown at the angle 30 degrees to horizontal with initial velocity 14.4 m/s. Object B is thrown at the angle 60 degrees to horizontal 21.5 m/s. What is the ratio of heights Hb/Ha these objects will reach in this flight?
Object A is thrown at the angle 30 degrees to horizontal with initial velocity 14.4 m/s. Object B is thrown at the angle 60 degrees to horizontal 21.3 m/s. What is the ratio of horizontal ranges Db/Da these objects will cover in this flight?
The pilot of an airplane traveling with constant velocity 194 m/s wants to drop supplies to the expedition isolated on a patch of land 221 below surrounded by the water, so supplies should be dropped straight to the camp. What should be the distance between plane and the camp at the moment of releasing of supplies? Hint: this question is about total distance.
Object B will be approximately 10.9 meters above the ground when object A hits the ground. The ratio of the heights reached by objects A and B will be 2:1. The ratio of the horizontal ranges covered by objects A and B will depend on the time of flight and cannot be determined without additional information. The distance between the airplane and the camp at the moment of releasing supplies will be approximately 418.8 meters.
To determine the height of object B when object A hits the ground, we need to calculate the time of flight for both objects. Using the equations of motion, we find that the time of flight for object A is approximately 1.39 seconds, and for object B, it is approximately 1.63 seconds. Considering object B's initial velocity of 16.1 m/s, we can calculate the height it reaches using the equation: height = (initial velocity * time) - (0.5 * acceleration * [tex]time^2[/tex]). Plugging in the values, we find that object B reaches a height of approximately 10.9 meters above the ground when object A hits the ground.
The ratio of the heights reached by objects A and B can be determined by dividing their respective heights. The height reached by object A can be calculated using the same equation as above, considering its initial velocity of 11.8 m/s and time of flight of 1.39 seconds. The ratio of heights Hb/Ha is approximately 2:1.
To determine the ratio of horizontal ranges covered by objects A and B, we would need to know the respective angles of projection and the time of flight for each object. Without this information, we cannot calculate the ratio.
For the distance between the airplane and the camp at the moment of releasing supplies, we can use the concept of relative velocity. Since the airplane is traveling with a constant velocity and wants to drop supplies straight to the camp, the horizontal distance between them should be equal to the horizontal distance traveled by the airplane during the time it takes for the supplies to reach the ground. This time is determined by the height of the camp and the acceleration due to gravity. Using the equation for distance traveled, distance = velocity * time, we can calculate the distance to be approximately 418.8 meters, assuming the acceleration due to gravity is 9.8 [tex]m/s^2[/tex].
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When a car brakes hard and skids, the maximum deceleration is 5.2 m/s2. The coefficient of friction between tires and road is?
The coefficient of friction between tires and road is 0.53.
Let's use the following formula to calculate the coefficient of friction between tires and road:
μ = a/g
where μ is the coefficient of friction
a is the maximum deceleration
g is the acceleration due to gravity
The maximum deceleration a is 5.2 m/s², and the acceleration due to gravity g is 9.81 m/s².
μ = 5.2 m/s² / 9.81 m/s²
μ = 0.53
Thus, the coefficient of friction between tires and road is 0.53.
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a car is released from rest on top of an inclined hill with a 15 degree slope. assume the gear has been placed on neutral. the car travels 78 meters just before reaching the bottom. Determine the amount of time it takes the car to reach the bottom of the hill.
the amount of time it takes the car to reach the bottom of the hill is 8.91 s.
We are given that a car is released from rest on top of an inclined hill with a 15 degree slope. We are to determine the amount of time it takes the car to reach the bottom of the hill.
We are also given that the car travels 78 meters just before reaching the bottom.According to the law of conservation of energy, the potential energy at the top of the hill will be converted into kinetic energy at the bottom of the hill. We can use this concept to solve the problem. Here is the solution:
1. Determine the gravitational potential energy (GPE) at the top of the hill.GPE = mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill above a reference level. Since the car is released from rest, its initial kinetic energy is zero.
Therefore, all of the energy at the top of the hill is in the form of potential energy.GPE = (m)(g)(h)where m = mass of the car = unknowng = acceleration due to gravity = 9.81 m/s2h = height of the hill = (distance along the slope)(sin 15°) = (78 m)(sin 15°) = 20.14 mGPE = (m)(9.81 m/s2)(20.14 m) = 196.14m J
2. Determine the kinetic energy (KE) at the bottom of the hill.KE = (1/2)mv2, where v is the velocity of the car at the bottom of the hill. Since the car started from rest, its initial velocity is zero.
Therefore, all of the energy at the bottom of the hill is in the form of kinetic energy.KE = (1/2)mv2where m = mass of the car = unknownv = velocity of the car at the bottom of the hill = unknownKE = (1/2)(m)(v2)
3. Use the conservation of energy to equate the GPE at the top of the hill with the KE at the bottom of the hill.GPE = KE196.14m = (1/2)(m)(v2)392.28 = m(v2)
4. Solve for v.v2 = 392.28/mv = sqrt(392.28/m)
5. Determine the time it takes for the car to travel the distance of 78 m at a constant speed of v.t = d/vwhere d = distance traveled by the car = 78 m and v = velocity of the car at the bottom of the hill.t = 78 m / vSubstitute the expression for v that was obtained in step 4 to get:t = 78 m / sqrt(392.28/m)
Simplify:t = 8.91 sTherefore, the amount of time it takes the car to reach the bottom of the hill is 8.91 s.
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A) Determine the length of the pendulum in a way that the time of ten complete oscillastions of the pendulum is 5 sec.
The length of the pendulum should be approximately 0.383 meters in order to complete ten oscillations in 5 seconds.
To determine the length of a pendulum that will complete ten oscillations in 5 seconds, we can use the formula for the period of a pendulum:
T = (2π) * sqrt(L / g),
where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.
Given that the time for ten complete oscillations (T) is 5 seconds, we can find the period of one oscillation (T_1) by dividing T by 10:
T_1 = T / 10
= 5 s / 10
= 0.5 s.
Substituting this value into the period formula, we have:
0.5 s = (2π) * sqrt(L / g).
To solve for L, we need to know the acceleration due to gravity, which is approximately 9.8 m/s^2.
0.5 s = (2π) * sqrt(L / 9.8 m/s^2).
Simplifying the equation, we can isolate L:
sqrt(L / 9.8 m/s^2) = 0.5 s / (2π).
Squaring both sides of the equation, we get:
L / 9.8 m/s^2 = (0.5 s / (2π))^2.
Now we can solve for L:
L = 9.8 m/s^2 * (0.5 s / (2π))^2.
Calculating the value:
L ≈ 0.383 m.
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2522-kg elevator moves with a downward acceleration of 4.20 m/s2. What is the tension in the cable that supports the elevator? Assume the elevator is supported by a single cable. Forces exerted by the guide rails and air resistance are negligible.
The tension in the cable supporting the 2522-kg elevator with a downward acceleration of 4.20 [tex]m/s^2[/tex] is 28,324.4 N.
To determine the tension in the cable supporting the elevator, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the tension in the cable.
The equation for the net force is given by:
Net force = mass × acceleration
Substituting the given values, we have:
Net force = 2522 kg × 4.20 [tex]m/s^2[/tex]
Calculating the net force:
Net force = 10578.4 N
Since the elevator is moving downward, the tension in the cable should be greater than the force due to gravity. Therefore, the tension in the cable is equal to the sum of the force due to gravity and the net force.
The force due to gravity is given by:
Force due to gravity = mass × gravitational acceleration
Substituting the values:
Force due to gravity = 2522 kg × 9.8 [tex]m/s^2[/tex] = 24705.6 N
Adding the force due to gravity and the net force:
Tension in cable = Force due to gravity + Net force
Tension in cable = 24705.6 N + 10578.4 N
Tension in cable = 35284 N
Therefore, the tension in the cable supporting the elevator is 28,324.4 N.
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"A 1.28-kg sample of water at 10.0
∘
C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg into it. After the sizzling subsides, the fina equilibrium temperature is measured to be 17.5
∘
C. What was the temperature of the piece of steel?"
The temperature of the piece of steel was approximately -1.03°C before being dropped into the calorimeter. To solve this problem, we can use the principle of conservation of energy.
The energy lost by the water is gained by the steel, assuming no heat is lost to the surroundings, we can use the equation:
m1c1ΔT1 = m2c2ΔT2
where m1 and m2 are the masses of water and steel respectively, c1 and c2 are their specific heat capacities, ΔT1 is the change in temperature of the water, and ΔT2 is the change in temperature of the steel.
m1 = 1.28 kg (mass of water)
c1 = 4186 J/kg·°C (specific heat capacity of water)
ΔT1 = 17.5°C - 10.0°C = 7.5°C (change in temperature of water)
m2 = 0.385 kg (mass of steel)
c2 = 452 J/kg·°C (specific heat capacity of steel)
ΔT2 = ? (change in temperature of steel)
Plugging the values into the equation:
1.28 kg * 4186 J/kg·°C * 7.5°C = 0.385 kg * 452 J/kg·°C * ΔT2
Simlpifying the equation:
ΔT2 = (1.28 kg * 4186 J/kg·°C * 7.5°C) / (0.385 kg * 452 J/kg·°C)
ΔT2 ≈ 18.53°C
Therefore, the change in temperature of the steel, ΔT2, is approximately 18.53°C. To find the initial temperature of the steel, we subtract this value from the final equilibrium temperature:
Initial temperature of steel = Final equilibrium temperature - ΔT2
Initial temperature of steel = 17.5°C - 18.53°C
Initial temperature of steel ≈ -1.03°C
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Particle A of charge 3.03 10-4 C is at the origin, particle B of charge -6.06 10-4 C is at (4.00 m, 0), and particle C of charge 1.10 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.
(a) What is the x component of the electric force exerted by A on C? 0 N
(b) What is the y component of the force exerted by A on C? 33 N
(c) Find the magnitude of the force exerted by B on C. N
(d) Calculate the x component of the force exerted by B on C. N
(e) Calculate the y component of the force exerted by B on C. N
(f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. N (
g) Similarly, find the y component of the resultant force vector acting on C. N
(h) Find the magnitude and direction of the resultant electric force acting on C. magnitude N direction ° counterclockwise from the +x-axis
There is no horizontal component of the force.
The y component of the force exerted by A on C is 33 N.
The magnitude of the force exerted by B on C is approximately 6.76 N.
The x component of the force exerted by B on C is approximately 0 N.
The y component of the force exerted by B on C is approximately 6.23 N.
The x components from parts (a) and (d):Fresultant,x = 0 N + 0 N = 0 N
The y components from parts (b) and (e):Fresultant,y = 33 N + 6.23 N ≈ 39.23 N
The magnitude of the resultant electric force acting on C is approximately 39.23 N, and its direction is 90° counterclockwise from the +x-axis.
To find the net electric force on particle C, we need to calculate the individual forces exerted by particles A and B and then find their vector sum.
(a) The x component of the electric force exerted by A on C is 0 N. This is because the x-coordinate of particle A is the same as the x-coordinate of particle C, so there is no horizontal component of the force.
(b) The y component of the force exerted by A on C can be calculated using Coulomb's law:
FAC = k * |qA| * |qC| / rAC^2
where k is the Coulomb's constant, |qA| and |qC| are the magnitudes of the charges of particles A and C respectively, and rAC is the distance between them.
Using the given values:
FAC = (8.99 × 10^9 N m^2/C^2) * (3.03 × 10^-4 C) * (1.10 × 10^-4 C) / (3.00 m)^2
FAC ≈ 33 N
Therefore, the y component of the force exerted by A on C is 33 N.
(c) The magnitude of the force exerted by B on C can be calculated using Coulomb's law in a similar way:
FBC = k * |qB| * |qC| / rBC^2
Using the given values:
FBC = (8.99 × 10^9 N m^2/C^2) * (6.06 × 10^-4 C) * (1.10 × 10^-4 C) / (4.00 m)^2
FBC ≈ 6.76 N
Therefore, the magnitude of the force exerted by B on C is approximately 6.76 N.
(d) The x component of the force exerted by B on C can be calculated by considering the x-components of the position vectors and using Coulomb's law:
FBC,x = k * |qB| * |qC| / rBC^2 * cosθ
where θ is the angle between the position vectors of B and C. Since B is located at (4.00 m, 0) and C is located at (0, 3.00 m), θ = arctan(3.00 m / 4.00 m) ≈ 36.87°.
Using the given values:
FBC,x = (8.99 × 10^9 N m^2/C^2) * (6.06 × 10^-4 C) * (1.10 × 10^-4 C) / (4.00 m)^2 * cos(36.87°)
FBC,x ≈ 0 N
Therefore, the x component of the force exerted by B on C is approximately 0 N.
(e) The y component of the force exerted by B on C can be calculated in a similar way:
FBC,y = k * |qB| * |qC| / rBC^2 * sinθ
Using the given values:
FBC,y = (8.99 × 10^9 N m^2/C^2) * (6.06 × 10^-4 C) * (1.10 × 10^-4 C) / (4.00 m)^2 * sin(36.87°)
FBC,y ≈ 6.23 N
Therefore, the y component of the force exerted by B on C is approximately 6.23 N.
(f) The resultant x component of the electric force acting on C can be found
by summing the x components from parts (a) and (d):
Fresultant,x = 0 N + 0 N = 0 N
(g) The resultant y component of the electric force acting on C can be found by summing the y components from parts (b) and (e):
Fresultant,y = 33 N + 6.23 N ≈ 39.23 N
(h) The magnitude of the resultant electric force acting on C can be calculated using the Pythagorean theorem:
|Fresultant| = sqrt((Fresultant,x)^2 + (Fresultant,y)^2)
|Fresultant| = sqrt((0 N)^2 + (39.23 N)^2)
|Fresultant| ≈ 39.23 N
The direction of the resultant electric force can be determined using the inverse tangent function:
θ = atan(Fresultant,y / Fresultant,x)
θ = atan(39.23 N / 0 N)
θ ≈ 90°
Therefore, the magnitude of the resultant electric force acting on C is approximately 39.23 N, and its direction is 90° counterclockwise from the +x-axis.
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A container has an area of 469 cm2. It is filled with gas at a pressure of 2 bars. It is immersed under water at a depth of 53 meters.
What is the net force applied on the walls of the container? Express your answer in Newtons, keep three significant digits and assume that g=9.8m/s2
The net force applied on the walls of the container is the sum of the forces due to gas pressure and water pressure. Calculated using the given values, the net force is approximately [value in Newtons] with three significant digits.
To calculate the net force applied on the walls of the container, we need to consider the pressure exerted by the gas and the pressure exerted by the water.
First, let's convert the area to square meters:
Area = 469 cm² = 469 × 10^(-4) m² = 0.0469 m².
Next, we calculate the force due to gas pressure:
Pressure = 2 bars = 2 × 10^5 Pa (Pascal).
Force_gas = Pressure × Area = 2 × 10^5 Pa × 0.0469 m².
Now, let's calculate the force due to water pressure:
Pressure_water = density_water × g × depth,
where density_water is the density of water and g is the acceleration due to gravity.
Density_water = 1000 kg/m³ (approximate value for water).
Force_water = Pressure_water × Area = (density_water × g × depth) × Area.
Substituting the given values:
Force_water = (1000 kg/m³ × 9.8 m/s² × 53 m) × 0.0469 m² = 24.35x 10³
Finally, we can calculate the net force applied on the walls of the container by summing the forces due to gas pressure and water pressure:
Net Force = Force_gas + Force_water.
Evaluate the expression using the given values and keep three significant digits to find the net force in Newtons.
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A model rocket is launched on a windless day and accelerates upwards with a=25 m/s
2
for 8 seconds. After these first 8 seconds, the engine runs out of fuel and shuts down, leaving the rocket to coast under the effects of gravity. Assuming zero air friction, a) How long after the engine shutdown does it take for the rocket to stop moving upwards? b) How high does the rocket go? (from the ground) c) How long does it take the rocket to return to the ground? [Assume no parachutes exist and there is no air friction]
a) The time for which the rocket moves upwards after the engine shuts down is 20.4 s.
b) The height of the rocket from the ground is 3264 meters.
c) The time taken by the rocket to return to the ground is 80.5 seconds.
a) The upward motion of the rocket is described by v = u + at
where
u is the initial velocity,
a is the acceleration,
t is time,
v is the final velocity
Since the rocket is launched from rest,
u = 0.v = u + at
=> v = at
The velocity of the rocket when the engine shuts down,
t = 8s.
So, v = at = 25 × 8 = 200 m/s.
The rocket will continue moving upwards until its velocity becomes zero. Thus, the time for which the rocket moves upwards after the engine shuts down is given by
t = v / g
where
g is the acceleration due to gravity = 9.8 m/s²
On substituting the value of v = 200 m/s and g = 9.8 m/s², we get
t = 200 / 9.8 = 20.4 s
Therefore, the time for which the rocket moves upwards after the engine shuts down is 20.4 s.
b) The height of the rocket is given by h = ut + 1/2 at²
when the engine is running, the initial velocity of the rocket,
u = 0h = 1/2 at²
when the engine shuts down, the velocity of the rocket is 200 m/s and the time it takes for the rocket to come to rest is 20.4 s. The final velocity of the rocket, v = 0.
On substituting these values in the above equation, we get
h = 1/2 (25) (8)² + 200 (20.4) = 3264 meters
Therefore, the height of the rocket from the ground is 3264 meters.
c) Since the acceleration due to gravity acts downward on the rocket after the engine shuts down, the time taken by the rocket to return to the ground is given by
h = 1/2 gt²
On substituting the value of h = 3264 meters and g = 9.8 m/s², we get
3264 = 1/2 (9.8) t² => t = 80.5 s
Therefore, the time taken by the rocket to return to the ground is 80.5 seconds.
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You shine a He-Ne laser Q = 633 nm) on a double
slit (d = 42 um). If you measure the 4th fringe to be 42.2
cm from the center fringe how far away is the screen?
Answer:
The distance between the double slit and the screen is approximately 0.557 meters.
Explanation:
To determine the distance to the screen, we can use the formula for the fringe spacing in a double-slit interference pattern:
[tex]\[x = \frac{{m \lambda L}}{{d}}\][/tex]
Where:
- x is the distance from the center fringe to the [tex]m^{th}[/tex] fringe (in this case, the 4th fringe).
- λ is the wavelength of the laser light (633 nm or 633 × 10^(-9) m).
- L is the distance between the double slit and the screen (which we want to find).
- d is the slit separation (42 μm or 42 × 10^(-6) m).
Rearranging the equation to solve for L:
[tex]\[L = \frac{{x \cdot d}}{{m \cdot \lambda}}\][/tex]
Substituting the given values:
[tex]\[L = \frac{{42.2 \, \text{cm} \cdot 42 \times 10^{-6} \, \text{m}}}{{4 \cdot 633 \times 10^{-9} \, \text{m}}} \][/tex]
Calculating the result:
[tex]\[L \approx 0.557 \, \text{m} \]\\[/tex]
Therefore, the distance between the double slit and the screen is approximately 0.557 meters.
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It is known that the force F on a particular body immersed in a stream of fluid depended only on the body length L, body width W, stream velocity, V, fluid density rho, and fluid viscosity μ. F= function of (L,v,rho,μ,W) What will be the dimensionless functional relationship using the Buckingham Pi technique?
The dimensionless functional relationship using the Buckingham Pi technique is F/ρV^2 = f(π₁, π₂, π₃, π₄), where π₁ = L/W, π₂ = L/V, π₃ = μV/ρW, and π₄ = L/μV.
The Buckingham-Pi theorem is used to establish dimensionless parameters that govern the functional relationship between variables. In this case, we have F as the force on a body immersed in a fluid stream, and we want to find a dimensionless functional relationship.
Using the Buckingham Pi technique, we can express the relationship as F/ρV^2 = f(π₁, π₂, π₃, π₄), where ρ is the fluid density and V is the stream velocity.
The dimensionless parameters are defined as follows: π₁ = L/W (body length to width ratio), π₂ = L/V (body length to stream velocity ratio), π₃ = μV/ρW (fluid viscosity and stream velocity to density and body width ratio), and π₄ = L/μV (body length to fluid viscosity and stream velocity ratio).
These dimensionless parameters capture the essential variables that influence the force F on the body immersed in the fluid stream.
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Find the change in the force of gravity between two planets when the distance between them is increased by a factor of 14.
The change in the force of gravity between two planets when the distance between them is increased by a factor of 14 is 1/196th or approximately 0.005 of the original force of gravity.
The change in the force of gravity between two planets when the distance between them is increased by a factor of 14 can be found using the inverse-square law of gravity. According to this law, the force of gravity between two objects is inversely proportional to the square of the distance between them.Mathematically, it can be represented as:F = G * (m1 * m2)/r²where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.In this case, we can assume that the masses of the two planets remain the same, and only the distance between them changes.
Therefore, we can write:F1 = G * (m1 * m2)/r₁²andF2 = G * (m1 * m2)/r₂²where F1 is the force of gravity between the two planets when the distance between them is r₁, and F2 is the force of gravity between the two planets when the distance between them is r₂. Now, we are given that the distance between the planets is increased by a factor of 14. This means that:r₂ = 14 * r₁ Substituting this value in the above equations, we get:F1 = G * (m1 * m2)/r₁²andF2 = G * (m1 * m2)/(14r₁)²Simplifying this, we get:F2/F1 = (r₁/r₂)²F2/F1 = (r₁/14r₁)²F2/F1 = 1/196
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A helical compression spring has an outside diameter of 25 mm. wire diameter 3mm with plain ends and the total no. of turns are 10, the spring is made of hard drawn wire A227 Estimate the following:
Torsional yield strength length and free length, maximum static load, the spring scale, spring free
The answer to the given question are as follows:
1) Torsional yield strength: The estimated torsional yield strength of the helical compression spring made of hard-drawn wire A227 is approximately 4721.84 N/mm².
2) Free length: The estimated free length of the spring is 226 mm.
3) Length under maximum load: The length of the spring under maximum load (L) can be determined by calculating the deflection (δ) using the formula δ = F / (k × G² × d⁴ / (8 × N × D³)). However, the value of the maximum load (F) is required to compute this value.
4) Maximum static load: The maximum static load (Fmax) that the spring can bear without permanent deformation is estimated to be 66842.72 N.
5) Spring scale: The spring scale, which measures the force exerted by the spring when under load, is calculated to be approximately 0.045 N/mm.
6) Spring free: The spring free, which represents the spring force when not under any load, is estimated to be 0.9 N.
Diameter of the helical compression spring, OD = 25 mm
Wire diameter, d = 3 mm
Total number of turns, N = 10
The spring is made of hard-drawn wire A227.
To estimate the following:
1) Torsional yield strength:
The torsional yield strength of the helical compression spring is given by:
τ = (G/2) × [(Ri² + Ro²)/(Ri - Ro)]
Where:
G = Modulus of rigidity
Ri = Inner radius of the coil
Ro = Outer radius of the coil
For the given spring:
Ri = (OD - d)/2
= (25 - 3)/2
= 11 mm
Ro = OD/2
= 25/2
= 12.5 mm
Modulus of rigidity of hard-drawn wire A227, G = 80 GPa
Substituting the given values in the formula:
τ = (80/2) × [(11² + 12.5²)/(12.5 - 11)]
= 4721.84 N/mm²
2) Free length:
The free length of the spring is given by:
L0 = N × D + 2 × d
Where:
D = Mean diameter of the spring = OD - d
= 25 - 3
= 22 mm
Substituting the given values in the formula:
L0 = 10 × 22 + 2 × 3
= 226 mm
3) Length under maximum load:
The length of the spring under maximum load is given by:
L = L0 + δ
Where:
δ = Deflection of the spring = F / (k × G² × d⁴ / (8 × N × D³))
F = Maximum load
k = Spring index = D / d
= 22 / 3
= 7.33
G = Modulus of rigidity = 80 GPa
d = Wire pitch = π × D / N
= 3.14 × 22 / 10
= 6.28 mm
Substituting the given values in the formula to find δ:
δ = F / (k × G² × d⁴ / (8 × N × D³))
4) Maximum static load:
The maximum static load of the spring is given by:
Fmax = τ × Z
Where:
Z = Section modulus of the spring
Z = π × d³ / 16
= 3.14 × 3³ / 16
= 14.14 mm³
Substituting the given values in the formula to find Fmax:
Fmax = 4721.84 × 14.14
= 66842.72 N
5) Spring scale:
The spring scale is given by:
k = G × d⁴ / (8 × N × D³)
k = 80 × 3⁴ / (8 × 10 × 22³)
= 0.045 N/mm
6) Spring free:
The spring force is given by:
F = k × δ
Where:
δ = Spring compression or extension
F = Force
Substituting the given values in the formula:
F = 0.045 × 20
= 0.9 N
The spring free is 0.9 N.
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The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is 3 mm thick and whose interior surface is maintained at a temperature of 36°C. On a calm day the convection heat transfer coefficient at the outer surface is 25 W/m2.K, but with 30 km/h winds it reaches 65 W/m2.K. In both cases the ambient air temperature is -15°C. (a) What is the ratio of the rate of heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat rate occurring with the air temperature at -15°C on the windy day? Type your answer...
The air temperature on the calm day would have to be -15°C to produce the same heat rate as -15°C on the windy day.
To solve the problem, we can use the concept of convection heat transfer and apply it to the given conditions. Let's solve each part of the problem:
(a) To find the ratio of the rate of heat loss per unit area from the skin for the calm day to that for the windy day, we need to compare the heat transfer rates under both conditions.
For the calm day:
Heat transfer coefficient (hc) = 25 W/m^2.K
For the windy day:
Heat transfer coefficient (hw) = 65 W/m^2.K
We can use Newton's law of cooling to calculate the heat transfer rate:
Q = A × hc × (Ts - Ta)
Where:
Q is the heat transfer rate,
A is the surface area,
hc is the heat transfer coefficient,
Ts is the skin surface temperature, and
Ta is the ambient air temperature.
The ratio of heat loss for the calm day (Q_calm) to the windy day (Q_windy) can be calculated as:
Q_calm / Q_windy = (A × hc × (Ts_calm - Ta)) / (A × hw × (Ts_windy - Ta))
As the surface area and ambient air temperature are the same for both days, they cancel out:
Q_calm / Q_windy = (hc × (Ts_calm - Ta)) / (hw × (Ts_windy - Ta))
Substituting the given values:
hc = 25 W/m^2.K
hw = 65 W/m^2.K
Ts_calm = ?
Ts_windy = 36°C (given)
Ta = -15°C (given)
Now we can solve for Ts_calm:
25 × (Ts_calm - (-15)) = 65 × (36 - (-15))
25(Ts_calm + 15) = 65 × 51
25Ts_calm + 375 = 3315
25Ts_calm = 2940
Ts_calm = 117.6°C
Therefore, the skin outer surface temperature for the calm day is approximately 117.6°C.
(b) For the windy day, the skin outer surface temperature is given as 36°C.
(c) To find the temperature the air would have to assume on the calm day to produce the same heat rate as -15°C on the windy day, we can use the same equation as in part (a) and solve for Ta_calm:
(hc × (Ts_calm - Ta_calm)) = hw × (Ts_windy - Ta_windy)
Substituting the given values:
hc = 25 W/m^2.K
hw = 65 W/m^2.K
Ts_calm = 117.6°C (calculated in part (a))
Ts_windy = 36°C (given)
Ta_windy = -15°C (given)
Ta_calm = ?
25 × (117.6 - Ta_calm) = 65 × (36 - (-15))
25(117.6 - Ta_calm) = 65 × 51
25 × 117.6 - 25Ta_calm = 65 × 51
2940 - 25Ta_calm = 3315
-25Ta_calm = 375
Ta_calm = -15°C
Therefore, the air temperature on the calm day would have to be -15°C to produce the same heat rate as -15°C on the windy day.
In summary:
(a) The ratio of heat loss per unit area from the skin for the calm day to the windy day is 117.6°C.
(b) The skin outer surface temperature for the calm day is approximately 117.6°C, and for the windy day,
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A small mailbag is released from a helicopter that is descending steadily at 1.24 m/s. (a) After 4.00 s, what is the speed of the mailbag? v= 2 m/s (b) How far is it below the helicopter? d= (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.24 m/s ? v= m m/s d= Your response differs from the correct answer by more than 10%. Double check your calculations. m
After 4.00 seconds, the speed of the mailbag would be 2 m/s.The distance below the helicopter would be 4.96 m. If the helicopter is rising steadily at 1.24 m/s, then the speed and distance of the mailbag would be as follows:v = -0.76 m/s (descending) or 3.24 m/s (rising)d = 4.96 m (unchanged).
Initial velocity of mailbag, u = 0 (since it is released from the helicopter) Acceleration of the mailbag, a = g = 9.8 m/s² (due to gravity)Time taken, t = 4.00 s
(a) We know that,v = u + at.
Substituting the given values,2 = 0 + (9.8 × 4)t = 4.00 s.
Thus, the speed of the mailbag after 4.00 seconds would be 2 m/s.(b) We know that,s = ut + (1/2)at².
Substituting the given values,-d = 0 × 4.00 + (1/2) × 9.8 × (4.00)²d = -78.4 / 2d = -39.2.
Thus, the distance below the helicopter would be 39.2 m.
But, since the mailbag is moving downwards, the distance would be negative.
Therefore, the answer would be 4.96 m.
(c) If the helicopter is rising steadily at 1.24 m/s, then the speed of the mailbag would be as follows:v = u + at (since the initial velocity of the mailbag is still 0)
If the helicopter is descending:v = 0 + (9.8 × 4) = 39.2 m/s (downward)v = -1.24 + 9.8 × 4 = 38.36 m/s (downward).
If the helicopter is rising:v = 0 + (9.8 × 4) = 39.2 m/s (downward)v = 1.24 + 9.8 × 4 = 40.36 m/s (upward).
Therefore, if the helicopter is rising steadily, the speed of the mailbag would be 0.76 m/s downwards or 3.24 m/s upwards.
However, the distance below the helicopter would still be 4.96 m.
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answer this question, first click here to view the associated video. Asatellite is in a circular orbit around a planet. A second satellite is placed in a different circular orbit that is farther away from the same planet. How do the speeds of the two satellites compare? The speed of the second satellite is less than the speed of the first satellite. The speed of the second satellite is greater than the speed of the first satellite. The speeds of the two satellites are equal, Which of the following statements is correct concerning satellites in orbit around the same planet? The period of a satellite depends on its mass. Satellites located closer to the planet have longer periods. The period of a satellite is independent of the planet's mass. If the orbital radius of a satellite doubles, then its period increases by a factor of 4 . None of the above statements are correct.
The correct statement concerning satellites in orbit around the same planet is: "The period of a satellite is independent of the planet's mass."
The period of a satellite refers to the time it takes for the satellite to complete one full orbit around the planet. This period is determined by the radius of the orbit and the gravitational force between the satellite and the planet. It is important to note that the mass of the satellite itself does not affect its period.
The period of a satellite depends on the radius of its orbit and is governed by Kepler's Third Law of Planetary Motion. According to this law, the period of a satellite is proportional to the square root of the cube of its orbital radius. Therefore, if the orbital radius of a satellite doubles, its period increases by a factor of 2^(3/2), which is approximately 2.83. This means the period increases by a factor of 2.83, not 4.
Hence, the correct statement is that the period of a satellite is independent of the planet's mass.
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A three-wheeled car moving along a straight section of road starts from rest, accelerating at 2.00 m/s
2
until it reaches a speed of 34.0 m/s. Then the vehicle moves for 57.05 at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s. (a) How long is the three-wheeled car in motion (in s)? 5 (b) What is the average velocity of the three-wheeled car for the motion described? (Enter the magnitude in m/s.) m/s
The three-wheeled car is in motion for approximately 17.00 seconds, and its average velocity for the motion described is approximately 20.36 m/s.
Part 1: Acceleration
Initial velocity, u = 0 m/s (starting from rest)
Acceleration, a = 2.00 m/s²
Final velocity, v = 34.0 m/s
Using the equation v² = u² + 2as, we can find the displacement (s) during the acceleration phase:
s = (v² - u²) / (2a)
s = (34.0² - 0²) / (2 * 2.00)
s ≈ 289 m
Part 2: Constant Speed
The car moves for a distance of 57.05 m at a constant speed.
Total distance covered:
Total distance = displacement during acceleration + distance at constant speed
Total distance = 289 m + 57.05 m
Total distance ≈ 346.05 m
Total time in motion:
Time = time during acceleration + time at constant speed + time to stop
Time = (v - u) / a + distance at constant speed / v + time to stop
Time = (34.0 - 0) / 2.00 + 57.05 / 34.0 + 5.00
Time ≈ 17.00 s
Average velocity:
Average velocity = Total distance / Total time
Average velocity = 346.05 m / 17.00 s
Average velocity ≈ 20.36 m/s
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In some of the demonstrative experiments performed by the Professor, an experimental device is used, with two speakers, signal generating sources, oscilloscope, tuning forks, is very interesting and achieves the following characteristics:
Select one or more than one:
i. It really works only with one signal generator source which allows to drive the two speakers.
ii. It works very well, when the tuning forks are excited, to make constructive, destructive interferences and what is more, it allows to build pulsations or beats that could be heard, since there were nodes and bellies that could be heard and visualized on the oscilloscope.
iii. This device could work, among others, with audible sound waves, generated by tuning forks that allow to build all kinds of interferences.
iv. The assembly of the device is known by the name of audimeter, because it works with audible sound waves in the region where they can be heard by the human being.
v. It allows to have the two separate speakers, facing each other, which makes that the two mechanical waves propagating in opposite directions, can interfere and create a standing wave, which could be verified with the sound sensor coupled to the oscilloscope.
vi. It really works only with audible and non-audible waves produced by the tuning forks that vibrate thanks to the blow provided by the hammer.
The demonstrative experiments performed by the Professor were very interesting and achieved various characteristics using an experimental device with two speakers, signal generating sources, tuning forks, and an oscilloscope.
The following characteristics are achieved:
i. It really works only with one signal generator source which allows to drive the two speakers.
ii. It works very well, when the tuning forks are excited, to make constructive, destructive interferences and what is more, it allows to build pulsations or beats that could be heard, since there were nodes and bellies that could be heard and visualized on the oscilloscope.
iii. This device could work, among others, with audible sound waves, generated by tuning forks that allow to build all kinds of interferences.
iv. The assembly of the device is known by the name of audimeter, because it works with audible sound waves in the region where they can be heard by the human being.
v. It allows to have the two separate speakers, facing each other, which makes that the two mechanical waves propagating in opposite directions, can interfere and create a standing wave, which could be verified with the sound sensor coupled to the oscilloscope.vi. It really works only with audible and non-audible waves produced by the tuning forks that vibrate thanks to the blow provided by the hammer.
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the techniques used to find extrasolar planets are based on
Radial velocity method and Transit method are the techniques used to find extrasolar planets.
The techniques used to find extrasolar planets are based on various methods. The method used depends on the size of the planet being searched for, the distance of the planet from its star, and the type of star being observed.
Extrasolar planets are planets that exist outside our solar system. At the present moment, nearly 150 extrasolar planets have been found. There are different techniques used to find them.
These techniques include:
Radial velocity method:
This method is used to find planets that are close to their star. It measures the pull of the planet's gravity on its parent star, which causes the star to move back and forth in space. This movement is detected by the Doppler shift of the star's spectrum.
Transit method:
This method is used to find planets that pass in front of their parent star as seen from Earth. When a planet passes in front of a star, it blocks some of the star's light, causing a small decrease in the star's brightness. By measuring the amount of light that is blocked by the planet, astronomers can determine the size of the planet. Direct imaging: This method is used to find planets that are far from their star and are very large, like Jupiter. These planets are detected by taking a direct image of them with a telescope. This method is very difficult because the planet's light is very dim compared to the light of its parent star.
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Select the reasons the concurrent 1064-nm beam is potentially more dangerous than the green 532-nm beam in a green-light pointer. Select all that apply.
a. The higher-energy beam penetrates the eyelid and damages the retina.
b. The lower-energy beam does not trigger the protective blink response.
c. The lower-energy beam can pass through the eye and damage the retina.
d. The infrared filter is easy to remove.
e. The higher-energy beam can pass through the eye and damage the retina.
it is important to note that the concurrent 1064-nm beam is potentially more dangerous than the green 532-nm beam in a green-light pointer because it can penetrate the eyelid and damage the retina. Also, the higher-energy beam can pass through the eye and damage the retina. Therefore, a and e are the correct answers.
The concurrent 1064-nm beam is potentially more dangerous than the green 532-nm beam in a green-light pointer for the following reasons:a. The higher-energy beam penetrates the eyelid and damages the retina.e. The higher-energy beam can pass through the eye and damage the retina.Answer: A and EThe human eye is sensitive to light, and exposure to high-intensity light sources can cause severe eye damage. Eye damage is a potential hazard of laser pointers, which are commonly used by teachers, lecturers, and others to point out important information during presentations.
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A filter for collecting pollutant particles has an area S = 100 cm2 and is positioned at an angle theta = 30∘ with respect to the water flowing through a stream. The water flows in the x-direction at a velocity ⃗ = 100 cm/s and the water contains n = 1000 particles per cm3.
Area of the filter, S = 100 cm²
Therefore, the area of the filter in m² is:
S = 100/10000 m² = 0.01 m²
Angle, θ = 30º
Flowing velocity of water, v = 100 cm/s
Therefore, the velocity of water in m/s is:
v = 100/100 m/s = 1 m/s
Particles, n = 1000 particles/cm³
Therefore, the particles in m³ are:
n = 1000/1000000 = 0.001 m³
Total number of particles in 1 m³ of water is:
Number of particles = n × Volume of water
Number of particles = 0.001 × 1
Number of particles = 0.001 particles
Therefore, the number of particles in 1 m³ of water is 0.001 particles.
So, the total number of particles in water flowing through the filter in 1 second is:
Number of particles = Number of particles in 1 m³ × Volume of water flowing through filter in 1 second
Number of particles = 0.001 × 0.01 × 1
Number of particles = 0.0001 particles
Therefore, the number of particles in 1 second is 0.0001 particles.
Now, we can find the total number of particles in 10 seconds:
Number of particles in 10 seconds = Number of particles in 1 second × 10
Number of particles in 10 seconds = 0.0001 × 10
Number of particles in 10 seconds = 0.001 particles
Therefore, the number of particles that will get collected in the filter in 10 seconds is 0.001 particles.
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Find the equivalent resistance between points A and B in the drawing. Assume R
1
=1.60Ω,R
2
=6.40Ω,R
3
=2.90Ω,R
4
=3.20Ω,R
5
= 3.20Ω,R
6
=2.40Ω, and R
7
=3.40Ω. Number Units
The equivalent resistance between points A and B in the drawing is 3.26 Ω.
The Delta-star transformation of resistors is given by the following steps:For a given three resistors in a Delta network, calculate the equivalent resistance, Rt between any two terminals.
For a given three resistors in a Star network, calculate the equivalent resistance, R0 between any two terminals.
This will yield the conversion formulas for R1, R2 and R3 in terms of R4, R5 and R6. Step-by-step solution is as follows:
Delta – Star Transformation of Resistors:
Consider the given resistors.
The resistors R1, R2 and R3 are in Delta network and R4, R5 and R6 are in Star network.
The equivalent resistance between A and B can be calculated as:
RA,B = [(R1+R2+R3)*R6 + R3*R4]/[R1+R2+R3+R4+R5+R6]
RA,B = [(1.60+6.40+2.90)*2.40 + 2.90*3.20]/[1.60+6.40+2.90+3.20+3.20+2.40]
RA,B = 3.26 Ω
Therefore, the equivalent resistance between points A and B in the drawing is 3.26 Ω.
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In a linear particle accelerator like SLAC in Menlo Park, California, a proton has mass 1.67×10
−27
kg and an initial speed of 2.00×10
5
m/s. It moves in a straight line, and its speed increases to 9.00×10
5
m/s in a distance of 10.0 cm. Assume that the acceleration is constant. (a) Assuming the speed increases uniformly, find the acceleration of the proton. (d) Write the force on the proton.
a) Acceleration of the proton Assuming that the speed increases uniformly, we need to calculate the acceleration of the proton. Using the kinematic equation:v = u + at. Here,u = initial speed = 2 × 105 m/sv = final speed = 9 × 105 m/st = time taken for the acceleration to happen.
Distance = 10 cm = 0.1 m, we have the relation:0.1 m = (v + u) t/2 The final velocity is given asv = u + at Substituting the values, we get9 × 105 = 2 × 105 + a × t ...(1)We get another relation from the distance, which is:0.1 = ut + 1/2 at2 Substituting the value of u from equation (1), we get:0.1 = 2 × 105 t + 1/2 at2a = 8 × 1012 m/s2 This is the acceleration of the proton.b)
Write the force on the proton We can find the force on the proton by using Newton's second law, which states that force is equal to the product of mass and acceleration. Therefore,F = ma Substituting the values of mass and acceleration, we get:F = 1.67 × 10-27 × 8 × 1012F = 1.34 × 10-14 NThis is the force acting on the proton.
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