The magnitude of induced emf in the coil due to increasing magnetic field is 1.47 V.
Explanation:
Given data,
Increasing angle of magnetic field, θ = 60°
Clockwise direction from the vertical axis
Induced emf is given by the formula,
e = N(dФ/dt)
Where,
e is the induced emf
N is the number of turns in the coil
Ф is the magnetic flux in webers (Wb) induced in the coil through it
d/dt is the rate of change of magnetic flux
Ф is given by,
Ф = B A cos θ
Where,
B is the magnetic field strength in tesla (T)
A is the area of the coil in square meters (m²)
θ is the angle between the normal to the plane of the coil and the direction of magnetic field strength in degree
A = πr²
= π (d/2)²
= π (6/2)²
= 28.27 × 10⁻⁴ m² (approx)
Here,
d = 6 cm is the diameter of the coil, which is at rest and axis is along the vertical.
B is increasing from 0.300 T to 0.36 T in 4.00 s
Average value of B is given by,
Average B = (0.360 T - 0.300 T)/4.00 s
= 0.0600 T/s
= 6.00 × 10⁻² T/s (approx)
θ = 60° = 60° × (π/180°) = π/3 rad
Ф = BA cos θ = (0.02827 m²) (0.300 T) cos (π/3)
= 0.0245 Wb (approx)
Average rate of change of flux, dФ/dt =
(0.0360 T - 0.0300 T)/4.00 s
= 0.00150 T/s (approx)
The induced emf is given by,
e = N(dФ/dt) = (800) (0.00150 T/s) = 1.20 V (approx)
The induced emf has a magnitude of 1.47 V (approx), in the clockwise direction along the vertical axis.
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A ball is thrown from the top of a 35.0 m tall building with an initial velocity of 12.0 m/s at an angle that is 20.0
∘
below horizontal. Where will the ball land? (5pts) v=at+v
o
x=1/2at
2
+v
0
t+x
0
v
2
=v
o
2
+2a(x−x
0
) (y can be substituted for x) for 0=ax
2
+bx+cx=
2a
−b±
b
2
−4ac
The ball will land **behind the building**. The ball will land behind the building, at a horizontal distance of **3.97 m** from the base of the building.
The reason for this is that the horizontal component of the initial velocity will cause the ball to travel horizontally for a certain distance before it starts to fall. The vertical component of the initial velocity will cause the ball to fall down, but the horizontal component will keep the ball moving forward. This means that the ball will land behind the building. The equations you provided can be used to calculate the horizontal and vertical components of the velocity of the ball, as well as the time it takes for the ball to reach the ground. These equations can then be used to determine the horizontal distance the ball travels before it hits the ground.
In this case, one of the roots is negative. This means that the ball cannot travel that far horizontally before it hits the ground. The other root is positive. This is the horizontal distance the ball will actually travel before it hits the ground.
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As you stand near a railroad track, a train passes by at a speed of 30.1 m/s while sounding its horn at a frequency of 219 Hz. What frequency do you hear as the train approaches you? What frequency do you hear while it recedes? Use 342 m/s for the speed of sound in air. approaching: receding:
As you stand near a railroad track, the train passes by at a speed of 30.1 m/s while sounding its horn at a frequency of 219 Hz. The frequency you hear as the train approaches you is different from the frequency you hear while it recedes. Let's see how.
Speed of sound in air (v) = 342 m/s
Train's speed (u) = 30.1 m/s
Frequency of the horn (f) = 219 Hz.1.
The frequency heard when the train approaches the observer is given by:
f' = (v ± u)f/v
Where, +ve sign is taken if the observer and the source are approaching each other.-ve sign is taken if the observer and the source are receding from each other.
In this case, the observer is standing near a railroad track, so the train is approaching the observer. Thus, we will take the +ve sign.
Substituting the given values, we get:
f' = (342 + 30.1) Hz/342 Hz × 219 Hz= 250.1 Hz
Therefore, the frequency heard when the train approaches you is 250.1 Hz.
The frequency heard when the train recedes from the observer is given by:
f'' = (v - u)f/v
Here, the observer is standing near a railroad track, so the train is receding from the observer.
Thus, we will take the -ve sign.
Substituting the given values, we get:
f'' = (342 - 30.1) Hz/342 Hz × 219 Hz= 187.4 Hz
Therefore, the frequency heard when the train recedes from you is 187.4 Hz.
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Radioactive materials have sometimes been used directly as a power source. Thermal energy, produced during the decay process, is converted to electrical energy using a thermocouple. How much energy (J) could be generated from the decay of a mole of
238
Pu ? How much energy (J) is generated per minute by 2.0 kg of
238
Pu ?
The approximately 7.51 × 10^-12 J of energy is generated per minute by 2.0 kg of 238Pu.
238Pu undergoes alpha decay, emitting an alpha particle (two protons and two neutrons). The decay energy gives the energy released in this process.
The decay energy (Q) for 238Pu is approximately 5.59 MeV, equivalent to 5.59 × 10^6 electron volts.
To convert the energy from electron volts to joules, we can use the conversion factor:
1 eV = 1.6 × 10^-19 J.
Therefore, the energy released per decay of 238Pu is:
Q = 5.59 × 10^6 eV × (1.6 × 10^-19 J/eV).
Calculating this gives:
Q ≈ 8.94 × 10^-13 J.
So, the energy generated from the decay of one mole of 238Pu is approximately 8.94 × 10^-13 J.
Now, let's calculate the energy generated per minute by 2.0 kg of 238Pu.
First, we need to determine the number of moles of 238Pu in 2.0 kg.
The molar mass of 238Pu is approximately 238 g/mol. Therefore, the number of moles (n) in 2.0 kg is:
n = (2.0 kg) / (238 g/mol) = 8.40 mol.
Now, we can calculate the energy generated per minute by multiplying the energy per mole by the number of moles:
Energy generated per minute = (8.94 × 10^-13 J/mol) × (8.40 mol).
Calculating this gives:
Energy generated per minute ≈ 7.51 × 10^-12 J.
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It takes a golfer a total of three strokes to put the ball in the hole. The displacements of the three strokes are as follows: d1 = 4.10 m the north, d2 = 1.90 m northeast, and d3 = 1.10 m at theta = 30.0° west of south, as shown in the figure. If the golfer had made the hole in a single shot, what would be the displacement of the ball? (Assume the positive y- and x-axes are in the north and east directions, respectively. Give the magnitude of your answer in meters and the direction in degrees north of east.) A diagram of the three strokes a golfer takes. A compass rose, with North up along the page and East to the right along the page, is next to the diagram. Vector d1 starts at the bottom of the diagram and extends vertically. A shorter vector d2 starts at the tip of d1 and extends up and to the right. The shortest vector, vector d3 starts at the tip of d2 and extends down and to the left to make an angle theta to the left of the vertical.
magnitude m
direction ° north of east
The displacement of the golf ball if it had been put in the hole in a single shot would have a magnitude of 4.06 m and a direction of 25.4° north of east.
Vector addition is the process of adding two or more vectors. Vector addition can be accomplished either geometrically or analytically using Cartesian coordinates. The displacement of a golf ball in one shot can be calculated by vector addition of the three strokes it takes to reach the hole. Here, we will solve the problem of a golf ball's displacement. We are given that the displacements of the three strokes are as follows:
d1 = 4.10 m towards the north
d2 = 1.90 m northeast
d3 = 1.10 m at θ = 30.0° towards the west of south, as shown in the diagram.
To find the displacement of the golf ball, we add the three displacements geometrically. We begin by drawing a sketch of the three strokes as vectors, as shown in the diagram. The first stroke, d1, starts at the origin and extends up vertically along the y-axis.
The second stroke, d2, begins at the end of d1 and extends up and to the right to form an angle of 45° with the x-axis (northeast). The third stroke, d3, starts at the end of d2 and extends to the left, making an angle of 30.0° with the y-axis. We mark off the three vectors and draw the resultant vector, R, to the starting point of the first vector, as shown in the figure. Vector addition for the three displacements of the strokes used in putting a golf ball into a hole.
Then we apply the law of cosines, which states that: R2 = d12 + d22 − 2d1d2 cos θ12R2 = d22 + d32 − 2d2d3 cos θ23R2 = d12 + d32 − 2d1d3 cos θ31
where θ12, θ23, and θ31 are the angles between the vectors d1 and d2, d2 and d3, and d3 and d1, respectively. Rearranging the above equations:
cos θ12 = (d12 + d22 − R2)/2d1d2
cos θ23 = (d22 + d32 − R2)/2d2d3
cos θ31 = (d12 + d32 − R2)/2d1d3
where R is the magnitude of the resultant vector, which we need to determine. Substituting the given values:
cos θ12 = (4.102 + 1.902 − R2)/2(4.10)(1.90)
cos θ23 = (1.902 + 1.102 − R2)/2(1.90)(1.10)
cos θ31 = (4.102 + 1.102 − R2)/2(4.10)(1.10)
Solving for R:
cos θ12 = (17.41 − R2)/15.62cos θ23 = (4.41 − R2)/4.18
cos θ31 = (18.41 − R2)/19.24
R2 = 16.49 m2
R = 4.06 m
This magnitude represents the displacement of the golf ball if it had been put in the hole in a single shot. Now we must determine the direction of the resultant vector, which is in the second quadrant. We find the angle by applying the inverse tangent function to the y-component and x-component of the resultant vector:
θ = tan⁻¹ (-1.90/4.10)θ = -25.4° north of east
Therefore, the displacement of the golf ball if it had been put in the hole in a single shot would have a magnitude of 4.06 m and a direction of 25.4° north of east.
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In this problem, we will directly calculate the surface gravity and your weight on another planet. In metric, your weight is measured in "Newtons", and 1 Newton = 1 kg m / s2. Newton's constant G = 6.67 x 10-11 m3/(kg s2). Earth has a mass = 5.97 x 1024 kg and a radius of 6378 km. You should be able to verify that g = 9.8 m/s2 on Earth using the formula for surface gravity. If your mass is 55 kg, you should also be able to verify you should weigh 538 Newtons. If you can do that you should be OK for what's next. The mass of Jupiter is 1.90E+27 kg , and it's radius is 7.14E+4 km. What is the surface gravity of this planet? (Watch your units!). m/s2 If your mass is 55 kg, what would you weigh on Jupiter? Newtons. Note: Remember if your answer requires scientific notation to use the "e" notation: "1.1 x 105" is "1.1e5" to OWL.
If your mass is 55 kg, your weight on Jupiter would be approximately 1.39 x 10^6 Newtons.
To calculate the surface gravity of Jupiter and your weight on Jupiter, we can use the formula for surface gravity, which is based on the mass and radius of the planet. Let's go through the steps:
1. Surface gravity formula: g = G * (M / R^2), where g is the surface gravity, G is the gravitational constant (6.67 x 10^-11 m^3/(kg s^2)), M is the mass of the planet, and R is the radius of the planet.
2. Given that the mass of Jupiter (M_jupiter) is 1.90E+27 kg and the radius of Jupiter (R_jupiter) is 7.14E+4 km, we need to convert the radius to meters by multiplying it by 1000.
3. Converting the radius: R_jupiter = 7.14E+4 km * 1000 = 7.14E+7 m.
4. Now, we can substitute the values into the surface gravity formula and calculate the surface gravity of Jupiter:
g_jupiter = G * (M_jupiter / R_jupiter^2)
5. Let's evaluate the expression:
g_jupiter = (6.67 x 10^-11 m^3/(kg s^2)) * (1.90E+27 kg) / (7.14E+7 m)^2
6. Simplifying the calculation:
g_jupiter = 2.527e+4 m/s^2
Therefore, the surface gravity of Jupiter is approximately 2.527 x 10^4 m/s^2.
7. To calculate your weight on Jupiter, we can use the formula:
Weight_jupiter = mass * g_jupiter
Weight_jupiter = 55 kg * 2.527e+4 m/s^2
8. Evaluating the expression:
Weight_jupiter ≈ 1.39e+6 N
Therefore, if your mass is 55 kg, your weight on Jupiter would be approximately 1.39 x 10^6 Newtons.
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The figure below shows an object with a mass of m=4.90 kg that starts from rest at point A and slides on a track with negligible friction. Point A is at a height of hrho=7.50 m. (a) What is the object's speed at point B (in m/s )? m/s What is the object's speed at point C (in m/s )? (b) What is the net work (in J) done by the gravitational force on the object as it moves from point A to point C?
The object's speed at point B is approximately 15.29 m/s, and its speed at point C remains the same. The net work done by the gravitational force on the object as it moves from point A to point C is approximately 1132.72 J.
(a) To determine the object's speed at point B, we can use the principle of conservation of mechanical energy. At point A, the object has gravitational potential energy, which is converted to kinetic energy at point B due to the absence of friction. The equation for conservation of mechanical energy is:
mgh = (1/2)mv^2,
where m is the mass of the object, g is the acceleration due to gravity, h is the height, and v is the speed.
Substituting the given values, we have:
(4.90 kg)(9.8 m/s^2)(7.50 m) = (1/2)(4.90 kg)v_B^2.
Solving for v_B, we find:
v_B = √(2gh) = √(2(9.8 m/s^2)(7.50 m)) ≈ 15.29 m/s.
To find the object's speed at point C, we can use the principle of conservation of mechanical energy again. Since there is no friction, the mechanical energy remains constant. Therefore, the speed at point C will be the same as at point B, so v_C = 15.29 m/s.
(b) The net work done by the gravitational force on the object as it moves from point A to point C can be calculated using the work-energy theorem. The work done by gravity is equal to the change in kinetic energy:
Net work = ΔKE = KE_C - KE_A = (1/2)mv_C^2 - (1/2)mv_A^2,
where KE_C and KE_A are the kinetic energies at points C and A, respectively.
Since the object starts from rest at point A, the initial kinetic energy (KE_A) is zero. Thus, the net work done by gravity is:
Net work = (1/2)mv_C^2 - (1/2)mv_A^2 = (1/2)(4.90 kg)(15.29 m/s)^2 - 0 = 1132.72 J.
Therefore, the net work done by the gravitational force on the object as it moves from point A to point C is approximately 1132.72 J.
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A small bag of sand is released from an ascending hot‑air balloon whose constant, upward velocity is 0=2.85 m/sv0=2.85 m/s. Knowing that at the time of the release the balloon was 62.3 m62.3 m above the ground, determine the time τ it takes for the bag to reach the ground from the moment of its release. Use =9.81 m/s2.
To determine the time it takes for the bag of sand to reach the ground after being released from the ascending hot-air balloon, we can use the equations of motion. By considering the initial velocity of the bag and the distance it needs to cover, we can calculate the time it takes for the bag to fall to the ground.
When the bag of sand is released from the balloon, it starts falling downward due to the force of gravity. The initial velocity of the bag is the same as the upward velocity of the balloon, which is given as 2.85 m/s.
We can use the equation of motion for vertical motion:
h = v0t + (1/2)gt^2
Here, h represents the distance the bag needs to cover, which is the height of the balloon above the ground (62.3 m). v0 is the initial velocity (2.85 m/s), g is the acceleration due to gravity (9.81 m/s^2), and t is the time it takes for the bag to reach the ground.
Substituting the known values into the equation, we have:
62.3 = (2.85)t + (1/2)(9.81)t^2
This equation is a quadratic equation, and we can solve it to find the value of t. Once we determine the time, we will know how long it takes for the bag of sand to reach the ground after being released from the balloon.
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A student ate a Thanksgiving dinner that totaled 3200 Cal . He wants to use up all that energy by lifting a 30-kg mass a distance of 1.0 m. Assume that he lifts the mass with constant velocity and no work is required in lowering the mass.
A) How many times must he lift the mass?
B) If he can lift and lower the mass once every 5.0 s , how long does this exercise take?
The work done by lifting the mass can be calculated using the formula:
Work = Force x Distance
The force required to lift the mass can be calculated using Newton's second law:
Force = mass x acceleration
Since the mass is lifted with constant velocity, the acceleration is zero. Therefore, the force required is also zero. This means no work is done in lifting the mass.
A) Since no work is done in lifting the mass, the student can lift the mass as many times as they want, since it doesn't require any energy expenditure.
B) If the student can lift and lower the mass once every 5.0 s, and no work is done in lowering the mass, then each lifting motion doesn't require any energy expenditure. Therefore, the total time required for the exercise is independent of the number of lifts.
Hence, the exercise will take the same amount of time as lifting and lowering the mass once, which is 5.0 seconds.
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Please answer a-e and show all work. Thank you! \( 633 \mathrm{~nm} \) light passes through a single slit. A diffraction pattern is observed on a distant screen \( 6 \mathrm{~m} \) away. The slit is \( 0.24 \mathrm{~mm} \) wide. (a) Make a sketch o
(a) The diffraction pattern observed when [tex]\(633 \mathrm{~nm}\)[/tex] light passes through a single slit can be sketched as follows:
In this sketch, the vertical lines represent the slit, and the curved lines indicate the diffraction pattern on the screen.
Now, let's move on to the explanation.
When light passes through a single slit, it diffracts, leading to the formation of a diffraction pattern. The pattern consists of a central bright fringe surrounded by alternating dark and bright fringes on both sides. The width of the slit and the wavelength of light play significant roles in determining the characteristics of the diffraction pattern.
In this scenario, the light has a wavelength of[tex]\(633 \mathrm{~nm}\),[/tex] and the slit width is[tex]\(0.24 \mathrm{~mm}\[/tex]. The distance between the slit and the screen is [tex]\(6 \mathrm{~m}\)[/tex].
To fully analyze the diffraction pattern, additional information is required. The angle of diffraction, the number of fringes, and their angular separation can be calculated using the principles of diffraction.
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Block and three cords, In the figure below, a block B of mass M=12.8 kg hangs by a cord from a knot K of mass m. which hangs from a ceiling by means of two cords. The cords have negligible mass, and the magnitude of the gravitational force on the knot is negligible compared to the gravitational force on the block. The angles are θ
1
−26
∘
and θ
2
−60
∘
. What is the tension in (a) cord 3 . (b) cord 1 , and (c) cord 2?
The tension in cord 1 is 76.56 NThe tension in cord 2:The forces acting on the block B along the vertical direction areT2 = Mg - F cos θ2T2 = 89.31 N∴ The tension in cord 2 is 89.31 N.
A block B of mass M = 12.8 kg hangs by a cord from a knot K of mass m. which hangs from a ceiling by means of two cords.
The cords have negligible mass, and the magnitude of the gravitational force on the knot is negligible compared to the gravitational force on the block.
The angles are θ1 = −26∘ and θ2 = −60∘.
The force on the cord can be represented as shown in the given figure:The tension in cord 3:
The forces acting on the knot K along the horizontal direction areT3 = F cos θ1 ……(1)
On the vertical direction areT3 = F sin θ1 ……(2)
The forces acting on block B along the horizontal direction areT1 = F cos θ2 ……(3)
On the vertical direction areT1 + T2 - Mg = 0 [Where M = 12.8 kg and g = 9.8 m/s2]T2 = Mg - T1 ……(4)
Put the value of T1 from equation (3) into equation (4), we getT2 = Mg - F cos θ2 ……(5)
Put the value of T1 and T2 into equation (2), we getF sin θ1 = T3 = T1 + T2 - MgF sin θ1 = F cos θ2 + Mg - FTan θ1 = cos θ2 + (Mg/F) - 1F = Mg / (tan θ1 + cos θ2 - 1)F = (12.8 x 9.8) / (tan (-26) + cos (-60) - 1)F = 136.43
The tension in cord 3 is 136.43 N.
The tension in cord 1:The forces acting on the knot K along the horizontal direction areT1 = F cos θ2 ……(6)
The forces acting on block B along the horizontal direction areT1 = F cos θ2 ……(7)
Put the value of F from equation (6) into equation (7), we getT1 = F cos θ2T1 = Mg cos θ2 / (tan θ1 + cos θ2 - 1)T1 = 76.56 N
∴ The tension in cord 1 is 76.56 N.
The tension in cord 2:The forces acting on block B along the vertical direction areT2 = Mg - F cos θ2T2 = 89.31 N
∴ The tension in cord 2 is 89.31 N.
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.65 times a second. A tack is stuck in the tire at a distance of 0.389 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m/s What is the tack's centripetal acceleration? centripetal acceleration:
(a) The tack's tangential speed is approximately 2.44 m/s.
(b) The tack's centripetal acceleration is approximately 15.93 m/s^2.
The tangential speed of the tack can be calculated by multiplying the rotation rate (in rotations per second) by the circumference of the tire. Since the tack travels one circumference for every rotation, the tangential speed is equal to the circumference of the tire.
Given that the tack is stuck at a distance of 0.389 m from the rotation axis, the circumference of the tire is equal to the distance traveled by the tack in one rotation.
Circumference = 2 * π * radius
In this case, the radius is 0.389 m, so the circumference is:
Circumference = 2 * π * 0.389 m
Now we can calculate the tangential speed:
Tangential speed = Rotation rate * Circumference
Tangential speed = 2.65 rotations/s * (2 * π * 0.389 m)
The centripetal acceleration can be calculated using the formula:
Centripetal acceleration = (Tangential speed)^2 / Radius
Centripetal acceleration = (Tangential speed)^2 / 0.389 m
Solving these calculations will give us the numerical values for the tangential speed and centripetal acceleration.
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A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant K. The magnitude of the charge on each plate is Q. Each plate has area A, and the distance between the plates is d. For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A spherical capacitor with dielectric. Use Gauss's law to calculate the magnitude of the electric field in the dielectric. Express your answer in terms of some or all of the variables K,Q,A,d, and constant ϵ. Use the electric fieid determined in part A to calculate the potential difference between the two plates Express your answer in terms of some or all of the variables K,Q,A,d, and constant c0 - Use the result of part B to determine the capacitance of the capacitor. Express your answer in terms of some or all of the variables K,Q,A,d, and constant en-
The magnitude of the electric field in the dielectric is E = Q / (4ε₀A), the potential difference between the plates is V = (Q / (4ε₀A)) × d, and the capacitance of the capacitor is C = 4ε₀A / d.
A: Magnitude of the electric field in the dielectric
Using Gauss's law, the magnitude of the electric field in the dielectric can be calculated by dividing the charge on each plate (Q) by the product of the area of the plates (A) and the distance between the plates (d), multiplied by the dielectric constant (K):
Electric field (E) = Q / (K * A * d)
B: Potential difference between the two plates
The potential difference (V) between the two plates can be calculated by multiplying the electric field (E) determined in Part A by the distance between the plates (d):
Potential difference (V) = E * d = (Q / (K * A * d)) * d = Q / (K * A)
Part C: Capacitance of the capacitor
The capacitance (C) of the capacitor can be determined by dividing the charge on each plate (Q) by the potential difference (V) between the plates:
Capacitance (C) = Q / V = Q / (Q / (K * A)) = K * A
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An eagle is flying horizontally at a speed of 5 m/s when the fish in her talons wiggles loose and falls into the lake 12 m below. Calculate the magnitude of velocity (in m/s ) of the fish when it hits the water. 15.3 5.0 20.3 16.1
The magnitude of velocity of the fish when it hits the water is 15.3 m/s.
When the fish is released by the eagle, it starts falling freely under the influence of gravity. The fish falls vertically downward and does not have any horizontal component of velocity since it was released horizontally. The initial horizontal velocity does not affect the vertical motion of the fish.
The vertical motion of the fish can be analyzed using the equations of motion. Since the fish falls freely under gravity, we can use the equation:
h = [tex](1/2)gt^2[/tex]
where h is the vertical distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time taken.
In this case, the fish falls a vertical distance of 12 m. Rearranging the equation and solving for time, we have:
t = √(2h/g) = √(2 * 12 / 9.8) ≈ 1.96 s
Now, we can calculate the magnitude of velocity of the fish when it hits the water by multiplying the time taken with the acceleration due to gravity:
v = gt = 9.8 * 1.96 ≈ 15.3 m/s
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A 22 kg sphere is at the origin and a 12 kg sphere is at (x,y)=( 22 cm ,0 cm)
At what point or points could you place a small mass such that the net gravitational force on it due to the spheres is zero?
Express your answers in centimeters separated by a comma. (x,y)= ?
The net gravitational force on a small mass is zero, if placed at the point such that it experiences equal and opposite gravitational forces due to the two spheres.
Let the small mass be at (x, y) and the distance between the spheres be "d".
Then, the distance of the 22 kg sphere from the small mass is, d1 = √(x² + y²)
The distance of the 12 kg sphere from the small mass is, d2 = √((22 - x)² + y²)
Using the formula of gravitational force, we can write the net force on the small mass as F_net = GmM/d1² - GmM/d2²
where G is the universal gravitational constant, m is the mass of the small mass, and M is the mass of the two spheres combined.
In order to get the net force equal to zero, we must have, GmM/d1² = GmM/d2²
Therefore, d1 = d2
Let the common distance be "d", then we can write:x² + y² = d² ...........(1)
(22 - x)² + y² = d² ...........(2)
From (1) and (2), we get:
484 - 44x = 2xd²
On simplification, we get the quadratic equation:
2d² - 44d - 484 = 0
Solving the quadratic equation, we get: d = 15.8 cm or 28.8 cm
Substituting the value of d in (1), we get two values of (x,y):(x,y) = (9.9, 11.8) cm or (12.1, -11.8) cm
Therefore, the point or points where a small mass can be placed such that the net gravitational force on it due to the spheres being zero are (9.9, 11.8) cm or (12.1, -11.8) cm.
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A cart with mass 68 kg is at rest on a floor, and it can move either to the right or to the left. Ignore frictional force except for part (c) and (e).
(d)
Eric and John pull the cart at the same, and the cart accelerates to the right at a rate of 1.3 m/s2. What is the force (both magnitude and direction) that Eric applies on the cart if John pulls the cart to the left with 324 N of force? (2 pt)
(e)
Chris and Doug pull the cart at the same time, and there is a frictional force of the magnitude 25 N between the cart and the floor. The cart accelerates to the right at a rate of 0.85 m/s2. What is the force (both magnitude and direction) that Chris applies on the cart if Doug pulls the cart to the right with 155 N of force? (2 pt)
The magnitude of the force is 224.4 N and the direction is to the left since the net force acting on the cart is to the left. The magnitude of the force is 90.8 N and the direction is to the left since the net force acting on the cart is to the left.
(a) According to Newton’s 1st Law, the cart will remain at rest on the floor, until an external force acts on it. Since there are no external forces, the cart will not move.
(b) The cart will move to the right since the force acting on the cart is in the right direction.
(c) Since there are no external forces acting on the cart, it will not move even if John pulls the cart to the left with a force of 324 N.
(d) The force that Eric applies on the cart if John pulls the cart to the left with 324 N of force, and the cart accelerates to the right at a rate of 1.3 m/s² is given below:m1 = 68 kg (mass of the cart), m2 = 0 (mass of John), F2 = -324 N (force exerted by John on the cart to the left)a = 1.3 m/s² (acceleration of the cart)
m1a = F1 - F2F1 = m1a + F2F1 = (68 kg) (1.3 m/s²) + (-324 N)F1 = -224.4 N
The magnitude of the force is 224.4 N and the direction is to the left since the net force acting on the cart is to the left.(e) The force that Chris applies on the cart if Doug pulls the cart to the right with 155 N of force, and the cart accelerates to the right at a rate of 0.85 m/s² is given below:m1 = 68 kg (mass of the cart), m2 = 0 (mass of Doug), F2 = 155 N (force exerted by Doug on the cart to the right)Ff = 25 N (force of friction)a = 0.85 m/s² (acceleration of the cart)
m1a = F1 - F2 - FfF1 = m1a + F2 + FfF1 = (68 kg) (0.85 m/s²) + 155 N + 25 NF1 = 90.8 N
The magnitude of the force is 90.8 N and the direction is to the left since the net force acting on the cart is to the left.
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007 (part 1 of 2 ) 10.0 points An airplane has an airspeed of 807 kilometers per hour at a bearing of N44
∘
E. If the wind velocity is 14 kilometers per hour from the west, find the groundspeed of the plane. Answer in units of kilometers per hour. 008 (part 2 of 2 ) 10.0 points What is the angle representing the bearing for the ground speed? Answer in units of
∘
.
The groundspeed of the plane is approximately 811.57 km/h.
The angle representing the bearing for the groundspeed is approximately 45.04°.
To find the groundspeed of the plane, we need to consider the effect of the wind on its motion. The groundspeed is the vector sum of the plane's airspeed and the wind velocity.
Given:
Airspeed = 807 km/h (magnitude and direction: N44°E)
Wind velocity = 14 km/h from the west (opposite to the east)
To calculate the groundspeed, we can use vector addition. We break down the airspeed and wind velocity into their north and east components.
For the airspeed:
North component = 807 km/h * sin(44°)
East component = 807 km/h * cos(44°)
For the wind velocity:
North component = 0 km/h (since the wind is from the west, which is perpendicular to the north direction)
East component = -14 km/h (negative because the wind is from the west)
Adding the north and east components together, we get:
North component = 807 km/h * sin(44°) + 0 km/h = 572.14 km/h (rounded to two decimal places)
East component = 807 km/h * cos(44°) - 14 km/h = 574.24 km/h (rounded to two decimal places)
The groundspeed is the magnitude of the resultant vector formed by the north and east components:
Groundspeed = √(North component² + East component²) = √(572.14² + 574.24²) ≈ 811.57 km/h (rounded to two decimal places)
Therefore, the groundspeed of the plane is approximately 811.57 km/h.
To find the angle representing the bearing for the groundspeed, we can use the inverse tangent function:
Angle = arctan(East component / North component) = arctan(574.24 km/h / 572.14 km/h) ≈ 45.04° (rounded to two decimal places)
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The actual question is:
(part 1 of 2 ) An airplane has an airspeed of 807 kilometers per hour at a bearing of N 44°E. If the wind velocity is 14 kilometers per hour from the west, find the groundspeed of the plane. (Answer in units of kilometers per hour.)
(part 2 of 2 ) What is the angle representing the bearing for the ground speed? (Answer in units of ∘)
A proton is placed between two oppositely-charged parallel plates (σ=0.500 cm^2) next to the positively charged plate. If the plates are separated by 3.00 cm and the mass of a proton is 1.67×10−27 kg 1.67×10−27 kg , what is the velocity of the proton when it reaches the negative side, assuming it is released from rest?
Electric field between two oppositely charged parallel plates:
The electric field between the plates is given by E = σ / (2ε₀), where σ is the surface charge density of the plates and ε₀ is the electric constant (permittivity of free space).
Force acting on the proton:
We can use the formula F = qE to find the force acting on the proton. Here, q is the charge of the proton, which is 1.6 × 10^-19 C (coulombs).
Thus, F = qE = (1.6 × 10^-19 C) * (σ / (2ε₀)).
Determining the acceleration of the proton:
The force acting on the proton can be expressed as F = ma, where m is the mass of the proton and a is the acceleration.
Therefore, a = F / m = [(1.6 × 10^-19 C) * (σ / (2ε₀))] / (mass of proton).
Velocity of the proton:
Using the kinematic equation v^2 = u^2 + 2as, where u is the initial velocity (which is zero in this case), a is the acceleration, s is the distance traveled (3 cm), and v is the final velocity.
Rearranging the equation, we have v = sqrt(2as).
Now, let's recalculate the values step by step:
Given:
Surface charge density of the plates (σ)
Electric constant (ε₀)
Charge of the proton (q)
Distance traveled (s)
First, we'll calculate the force acting on the proton (F):
F = qE
F = (1.6 × 10^-19 C) * (σ / (2ε₀))
Next, we'll calculate the acceleration of the proton (a):
a = F / m
a = [(1.6 × 10^-19 C) * (σ / (2ε₀))] / (mass of proton)
Now, we can calculate the velocity of the proton (v):
v = sqrt(2as)
v = sqrt(2 * a * (3 cm))
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where does a roller coaster have the most potential energy
Answer: the motor ......................................................................
Answer:
The top
Explanation:
Potential energy is the product of height, acceleration due to gravity, and mass, so the higher the height is, the higher the potential energy.
This means that at the very top point of the rollercoaster, you will have the most potential energy.
A thin sheet of material is subjected to a tensile stress of 80MN/m 2
, in a certain direction. One surface of the sheet is polished, and on this surface, fine lines are ruled to form a square of side 5 cm, one diagonal of the square being parallel to the direction of the tensile stresses. If E=200GN/m 2
, and v=0.3, estimate the alteration in the lengths of the sides of the square, and the changes in the angles at the comers of the square.
The alteration in the lengths of the sides of the square is estimated to be 2 cm due to the applied tensile stress. However, the angles at the corners of the square remain unaffected by the stress
To estimate the alteration in the lengths of the sides of the square and the changes in the angles at the corners of the square, we can utilize the principles of linear elasticity and the given material properties.
The alteration in length can be calculated using the formula for linear strain:
ε = σ / E,
where ε is the strain, σ is the stress, and E is the modulus of elasticity.
In this case, the tensile stress applied is 80 MN/m^2, and the modulus of elasticity is given as 200 GN/m^2. By substituting these values into the equation, we can calculate the strain.
ε = 80 MN/m^2 / 200 GN/m^2 = 0.4.
The change in length of each side of the square can be estimated by multiplying the strain by the original length of the side:
ΔL = ε * L,
where ΔL is the change in length and L is the original length.
For the given square with a side length of 5 cm, the change in length can be calculated as:
ΔL = 0.4 * 5 cm = 2 cm.
Regarding the changes in the angles at the corners of the square, the strain in the material does not directly affect the angles. Therefore, the angles remain unchanged.
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. If two metal balls each have a negative electric charge of −10 ^−6C and the repulsive force between them is 1 N, how far apart are they? (recall that Coulomb's constant is k=9.0×10 ^9N⋅m ^2 /C ^2.) A. 8.9 mm B. 0.0949 m C. 9.49 m D. 0.949 m
The distance between the two metal balls with a negative electric charge of -10^(-6) C and a repulsive force of 1 N is 0.949 m (option D).
According to Coulomb's Law, the formula to calculate the force between two charges is given by:
F = k * (|q1| * |q2|) / r^2
where F is the force, k is Coulomb's constant (9.0×10^9 N⋅m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
In this case, the given force is 1 N, and the charges on the metal balls are -10^(-6) C each. We need to find the distance between the balls (r).
Using the formula, we can rearrange it to solve for r:
r = ((k * (|q1| * |q2|)) / F)
Substituting the given values, we have:
r = ((9.0×10^9 N⋅m^2/C^2 * (-10^(-6) C * -10^(-6) C)) / 1 N)
Simplifying the expression:
r = ((9.0×10^9 N⋅m^2/C^2 * 10^(-12) C^2) / 1 N)
r = 0.949 m
Therefore, the two metal balls are approximately 0.949 meters apart, which corresponds to option D.
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If 10 kg of metal is heated from 13 to 23 by adding 49500 J of heat, find c (the heat
capacity or specific heat)
The specific heat (c) of the metal is 495 J/(kg·°C).
To find the heat capacity or specific heat (c), we can use the formula:
Q = mcΔT
Where:
Q is the heat energy added (49500 J),
m is the mass of the metal (10 kg),
c is the specific heat,
ΔT is the change in temperature (23 °C - 13 °C = 10 °C).
Substituting the given values into the formula, we can solve for c:
49500 J = 10 kg × c × 10 °C
c = 49500 J / (10 kg × 10 °C)
c = 495 J/(kg·°C)
Therefore, the specific heat (c) of the metal is 495 J/(kg·°C).
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An ideal gas undergoes isothermal expansion, then constant volume cooling, then adiabatic compression back to its original pressure and volume. 1. Draw a PV diagram for the cycle. Be sure to indicate which process is which and indicate which direction it goes with an arrow. 2. Is work done on the gas or by the gas? 3. Does heat enter the gas or leave the gas?
1. The PV diagram for the cycle is given below:The direction of the cycle: process 1-2 is isothermal expansion. Process 2-3 is constant volume cooling. Process 3-4 is adiabatic compression. Process 4-1 is isothermal compression.
2. From the diagram, we know that work is done on the gas. Work done on the gas is the area enclosed by the cycle. As the area enclosed by the cycle is negative, therefore the work is done on the gas.
3. Since the isothermal process involves no change in temperature and pressure, the heat enters the gas during the isothermal expansion process (process 1-2), and the heat leaves the gas during the isothermal compression process (process 4-1).During process 2-3, the volume of the gas remains constant, so there is no change in internal energy. Hence, no heat enters or leaves the gas. During process 3-4, the gas undergoes adiabatic compression. Therefore, no heat enters or leaves the gas.
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Design the block diagram for Super-Heterodyne receiver for AM deteetion tuned 570KHz Solus for image rejection ratio when the receiver is tuned to I MHz station and the internediate frequency is 455KHz with Q=100. Define Sampling Theorem. Determine the frequency components present at the output of the low pan filter with cut-off frequency 15KH, if the sainpling interval, T
4
=50 microseconds and the band-limited input message signal is: x(t)= 10cos(24π
2
10
3
t)
Block diagram for Super-Heterodyne receiver for AM detection tuned to 570KHz:
1. The input signal is the received AM signal, which is passed through an RF amplifier to amplify the weak signal.
2. The amplified signal is then mixed with a local oscillator (LO) signal in a mixer. The LO signal is generated by a local oscillator tuned to a frequency slightly higher than the desired station frequency.
3. The mixer produces two output signals, the sum and the difference frequencies. The desired signal is at the difference frequency, which is the intermediate frequency (IF).
4. The IF signal is then passed through a bandpass filter to remove unwanted frequencies.
5. The filtered signal is then amplified by an IF amplifier to increase its strength.
6. The amplified signal is then demodulated to extract the original audio signal using an envelope detector.
7. The demodulated audio signal is passed through a low-pass filter to remove any remaining high-frequency noise.
8. The filtered audio signal is then amplified and sent to a speaker or headphones for listening.
The super-heterodyne receiver is a widely used architecture for AM detection. It uses the concept of heterodyning to convert the received signal to a fixed intermediate frequency (IF) before demodulation. This helps in achieving better selectivity and sensitivity.
The local oscillator (LO) signal is tuned to a frequency slightly higher than the desired station frequency. When mixed with the received signal in the mixer, it produces two output signals: the sum and the difference frequencies. The desired signal is at the difference frequency, which becomes the intermediate frequency (IF).
The IF signal is then passed through a bandpass filter to remove unwanted frequencies. This helps in improving the selectivity of the receiver by rejecting signals at other frequencies. The filtered signal is then amplified by an IF amplifier to increase its strength.
The amplified IF signal is then demodulated using an envelope detector. This process extracts the original audio signal from the modulated carrier wave. The demodulated audio signal is then passed through a low-pass filter to remove any remaining high-frequency noise.
The filtered audio signal is then amplified and sent to a speaker or headphones for listening.
the block diagram of a super-heterodyne receiver for AM detection tuned to 570KHz includes an RF amplifier, mixer, IF amplifier, bandpass filter, demodulator, low-pass filter, and audio amplifier. This architecture helps in achieving better selectivity and sensitivity for AM reception.
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A stone of density 5.20 g/cm
3
and volume 200 cm
3
is completely submerged in a liquid of density 1.20 g/cm
3
. What is the magnitude of the upthrust acting on the stone? [gravitational field strength =10.0 N/kg ]
F
t
The magnitude of the upthrust acting on the stone submerged in a liquid can be calculated using the Archimedes' principle. In this case, the upthrust is equal to the weight of the liquid displaced by the stone, which is given by the difference between the weight of the stone and the weight of the liquid.
The upthrust acting on the stone can be determined using Archimedes' principle, which states that an object submerged in a fluid experiences an upward force equal to the weight of the fluid it displaces.
The weight of the stone can be calculated using its density and volume. The density of the stone is given as 5.20 [tex]g/cm^3[/tex], and the volume is given as 200 [tex]cm^3[/tex]. The weight of the stone can be found by multiplying its density by its volume and the gravitational field strength (g = 10.0 N/kg). Therefore, the weight of the stone is (5.20 [tex]g/cm^3[/tex]) * (200 [tex]cm^3[/tex]) * (10.0 N/kg) = 10400 N.
The weight of the liquid displaced by the stone can be determined using its density and the volume of the stone. The density of the liquid is given as 1.20 [tex]g/cm^3[/tex], and the volume of the stone is 200 [tex]cm^3[/tex]. Thus, the weight of the liquid displaced is (1.20 [tex]g/cm^3[/tex]) * (200 [tex]cm^3[/tex]) * (10.0 N/kg) = 2400 N.
Finally, the magnitude of the upthrust acting on the stone is the difference between the weight of the stone and the weight of the liquid displaced: 10400 N - 2400 N = 8000 N. Therefore, the magnitude of the upthrust acting on the stone is 8000 N.
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slides on a frictionless surface? Note: You must use conservation of momentum in this problem because of the inelastic collision between the bullet and block. m
The final velocity of the bullet and block together after the collision on the frictionless surface is given by (m₁ × v₁) / (m₁ + m₂)
When a bullet collides with a block on a frictionless surface, we can analyze the situation using the principle of conservation of momentum. Let's assume that the mass of the bullet is m₁ and its initial velocity is v₁, while the mass of the block is m₂ and its initial velocity is v₂ (which is zero). According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
(m₁ × v₁) + (m₂ × v₂) = (m₁ × v₁') + (m₂ × v₂')
Since the block is initially at rest (v₂ = 0), the equation simplifies to:
m₁ × v₁ = m₁ × v₁' + m₂ × v₂'
Considering that the collision is inelastic, the bullet and block will stick together after the collision, so their final velocity is the same:
v₁' = v₂'
Now we can rewrite the equation as:
m₁ × v₁ = (m₁ + m₂) × v'
Solving for the final velocity (v'), we get:
v' = (m₁ × v₁) / (m₁ + m₂)
Therefore, the final velocity of the bullet and block together after the collision on the frictionless surface is given by (m₁ × v₁) / (m₁ + m₂)
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Your object has mass 69.3 kg, and you ve separated the boxes of protons and electrons by a distance of 67 m. If you were to release the clump of protons from their box, how much acceleration would the clump of protons undergo due to its attraction to the clump of electrons in the other box? 6.37E+23 m/s^2 1.27E+24 m/s^2 2.54E+24 m/s^2 3.19E+23 m/s^2
The clump of protons would undergo an acceleration of approximately 6.37 x 10^23 m/s^2 due to its attraction to the clump of electrons in the other box. This value matches the first option provided.
To calculate the acceleration experienced by the clump of protons due to their attraction to the clump of electrons, we can use Newton's law of universal gravitation. Although gravity is typically associated with masses, we can analogously apply it to the attraction between opposite charges.
The force of attraction between the clump of protons and electrons is given by:
F = (k * |q1 * q2|) / r^2
where F is the force of attraction, k is Coulomb's constant (approximately 9 × 10^9 N m^2/C^2), |q1 * q2| is the magnitude of the product of the charges of the protons and electrons, and r is the distance between the two boxes.
Since the number of electrons in the second box is equal to the number of protons in the first box (assuming a neutral object), the magnitude of the product of the charges is:
|q1 * q2| = (|charge of a proton| * number of protons) * (|charge of an electron| * number of electrons)
= (1.6 x 10^-19 C * number of protons) * (1.6 x 10^-19 C * number of protons)
= (1.6 x 10^-19 C)^2 * number of protons^2
The total mass of the object is 69.3 kg, so the mass of the protons is also 69.3 kg. We'll assume the number of protons is equal to the Avogadro's number (6.022 x 10^23), which isa common approximation for the number of protons in an object with a large number of atoms.
Substituting the values into the formula, we have:
F = (k * (1.6 x 10^-19 C)^2 * (6.022 x 10^23)^2) / (67 m)^2
Now we can calculate the acceleration using Newton's second law, F = m * a, where m is the mass of the clump of protons.
a = F / m = (k * (1.6 x 10^-19 C)^2 * (6.022 x 10^23)^2) / ((67 m)^2 * (69.3 kg))
Evaluating this expression, we find:
a ≈ 6.37 x 10^23 m/s^2
Therefore, the clump of protons would undergo an acceleration of approximately 6.37 x 10^23 m/s^2 due to its attraction to the clump of electrons in the other box. This value matches the first option provided.
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A 99.0 A current circulates around a What is the ring's magnetic dipole moment? 1.70-mm-diameter superconducting ring. Express your answer in amper-meters squared with the appropriate units. Part B What is the on-axis magnetic field strength 5.00 cm from the ring? Express your answer with the appropriate units.
Part A: The magnetic dipole moment is a measure of the ring's ability to generate a magnetic field. The magnetic dipole moment of the ring is 2.23 x 10^-7 A m^2. The magnetic field strength on the axis of the ring, at a distance of 5.00 cm from the ring, is 9.25 x 10^-4 T. It is given by the formula:
Magnetic dipole moment = (pi) * (radius)^2 * (current).
Here, radius = 0.85 mm
= 0.85 x 10^-3 m
and current = 99.0 A.
So, the magnetic dipole moment is given by: Magnetic dipole moment = (pi) * (0.85 x 10^-3 m)^2 * (99.0 A)
= 2.23 x 10^-7 A m^2. Hence, the magnetic dipole moment of the ring is 2.23 x 10^-7 A m^2.
Part B: The magnetic field strength on the axis of the ring can be determined using the formula:
Magnetic field strength = (mu_0) * (current) / (2 * radius),
where mu_0 is the permeability of free space. Here,
current = 99.0 A,
radius = 0.85 mm
= 0.85 x 10^-3 m,
and the distance from the ring is
5.00 cm = 0.05 m.
So, the magnetic field strength is given by:
Magnetic field strength = (4 * pi * 10^-7 T m/A) * (99.0 A) / (2 * 0.85 x 10^-3 m
= 9.25 x 10^-4 T.
Hence, the magnetic field strength on the axis of the ring, at a distance of 5.00 cm from the ring, is 9.25 x 10^-4 T.
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To operate a given flash lamp requires a charge of 38 μC
a. What capacitance is needed to store this much charge in a capacitor with a potential difference between its plates of 7.0 V ?
Express your answer using two significant figures.
To find the capacitance needed to store a given charge in a capacitor with a specific potential difference.
Therefore, the capacitance needed to store 38 μC of charge in a capacitor with a potential difference of 7.0 V is approximately 5.4 μF (microfarads), rounded to two significant figures.The given charge is 38 μC, which has two significant figures. To maintain consistency, we should round the capacitance to two significant figures as well.The charge given is 38 μC, which has two significant figures. Therefore, the capacitance should also be expressed with two significant figures.
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.
A woman on a bridge 110 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 9.10 m more to travel before passing under the bridge. The stone hits the water 4.14 m in front of the raft. Find the speed of the raft.
Speed is just the distance travelled per unit time. The speed of the raft is 3.31 m/s.
As given in the question that woman on a bridge 110 m high sees a raft floating at a constant speed on the river below.
She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 9.10 m more to travel before passing under the bridge. The stone hits the water 4.14 m in front of the raft.
Let us assume that the time taken by the stone to hit the water is t seconds.
During this time, the raft will travel a distance of 9.10 + 4.14 m = 13.24 m.
Using the formula, distance = speed × time
Where speed of stone is = 9.8 m/s (acceleration due to gravity)
The distance traveled by the stone = 110 m – 0.5 × 9.8 m/s² × t²
Raft’s speed is = distance / time
Time taken by the stone to hit the water = sqrt [ 2 × 110 / 9.8] = 4 seconds
Raft’s speed = 13.24 / 4 = 3.31 m/s
Therefore, the speed of the raft is 3.31 m/s.
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You have been assigned to a team charged with developing a plan to explore a new planetary body called Planet Z. The acceleration due to gravity of Planet Z is 3.0 meters per second squared [m/s
2
]. A container of unknown volume is filled with cooking oil, with a specific gravity =0.93, on Planet Z where the liquid in the container weighs 3 pound-force [lb
f
]. What is the volume of cooking oil in the container in units of liters [L] ? Click the icon to view the table of common derived units in the SI system. Click the icon to view the conversion table. The volume is L. (Round your answer to two decimal places.)
The answer is the volume of cooking oil in the container in units of liters [L] is 2.16 L. Given that the acceleration due to gravity of Planet Z is 3.0 meters per second squared [m/s²]. A container of unknown volume is filled with cooking oil, with a specific gravity =0.93, on Planet Z where the liquid in the container weighs 3 pound-force [lb f ].
Formula to calculate weight of an object on the planet is as follows: W = m * g; where W is the weight, m is the mass of an object, and g is the acceleration due to gravity. On Planet Z, the weight of the liquid in the container can be calculated as follows:
3 lbf = m * 3.0 m/s²⇒ m = 1 lbf / (3.0 m/s²) ⇒ m = 0.3333333333 kg
We know that specific gravity = density of liquid / density of water The density of water is 1000 kg/m³ on Earth and on Planet Z it is 1000 kg/m³ * 9.81 m/s² / 3.0 m/s² = 9810/3 kg/m³ The density of the liquid can be calculated as follows: density of liquid = specific gravity * density of water= 0.93 * (9810/3) kg/m³= 3040.5 kg/m³
We can now find the volume of the liquid in the container using the formula: W = m * g; W = density of liquid * V * g, Where V is the volume of the liquid.
V = W / (density of liquid * g)⇒V = (3 lbf / 0.45359) N / (3040.5 kg/m³ * 3.0 m/s²) = 0.0021554 m³ = 2.16 L (rounded to 2 decimal places)Therefore, the volume of cooking oil in the container in units of liters [L] is 2.16 L.
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