The speed of the satellite is approximately 10,912 m/s.
The mass of the Earth is ME = 5.98 × 1024 kg.
The radius of the Earth is RE = 6.37 × 106 m.
The gravitational constant is G = 6.67 × 10−11 N·m2/kg2.The height of the satellite above the surface of the Earth is h = 847 km = 8.47 × 105 m.
The mass of the satellite is m = 6,754 kg.
The force of gravity that acts on the satellite Fg is equal to the centripetal force Fc.
The centripetal force is given by Fc = (mv²) / r where m is the mass of the object, v is its speed, and r is the radius of the circular path that the object is following.
The force of gravity is given by Fg = (GME m) / r², where ME is the mass of the Earth and r is the distance between the center of the Earth and the center of the satellite, which is equal to the sum of the radius of the Earth and the height of the satellite above the surface of the Earth.
Therefore, we have:(mv²) / r = (GME m) / r²We can simplify this equation by canceling out the m on both sides:(v²) / r = (GME) / r²Now we can solve for v:v² = (GME) / rv = √[(GME) / r]We can substitute the values given in the problem:
v = √[(6.67 × 10−11 N·m2/kg2 × 5.98 × 1024 kg) / (6.37 × 106 m + 8.47 × 105 m)]v = √(3.98676 × 1014 / 7.22 × 106)v = 10,912 m/s
Therefore, the speed of the satellite is approximately 10,912 m/s.
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Use source transformation to find iO in the given circuit where I=5 A. The current iO in the given circuit is A.
Source transformation is a technique in which the voltage source and current source of a circuit can be transformed into one another using the principle of electrical equivalence without changing the other electrical characteristics of the circuit.
When a voltage source is transformed into a current source, the internal resistance of the source and the load resistance of the circuit are changed accordingly. Similarly, when a current source is transformed into a voltage source, the internal resistance of the source and the load resistance of the circuit are also changed accordingly.
The given circuit is shown below: Where I = 5A and iO is the current in the circuit.First, we will transform the current source to a voltage source. This is done by multiplying the current source value (5A) by the resistance of the circuit (8Ω) to get the voltage source value.
Voltage source value = 5A x 8Ω = 40VThe equivalent circuit with the voltage source is shown below: Next, we can combine the two parallel resistors (4Ω and 8Ω) into one equivalent resistor (2.67Ω).The equivalent circuit is shown below: Now we can transform the voltage source back into a current source.
This is done by dividing the voltage source value (40V) by the equivalent resistance of the circuit (2.67Ω).Current source value = 40V / 2.67Ω = 15AThe direction of the current is opposite to the direction of the original current source.
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A canon fires off a cannonball at an angle of 38.0
∘
with a velocity of 135 m/s. Assuming negligible air resistance, how high above ground does the cannonball go, and how far has the cannonball traveled horizontally when it reaches its maximum height? (5 points) For the cannonball in problem 4, the magnitude of its velocity is greatest a) immediately after it is fired, b) at the highest point of its trajectory, c) somewhere between the lowest and highest points of its trajectory or d) the magnitude of the velocity is the same at all points in time.
The cannonball travels approximately 897.26 m horizontally when it reaches its maximum height. Regarding the additional question, the magnitude of the velocity is greatest immediately after it is fired.
To determine the height above the ground reached by the cannonball and the horizontal distance traveled when it reaches its maximum height, we can use the following equations:
Vertical motion equation: h = v₀y * t + (1/2) * a * t²
Horizontal motion equation: x = v₀x * t
Angle of projection (θ) = 38.0°
Initial velocity (v) = 135 m/s
Acceleration due to gravity (a) = 9.81 m/s²
First, let's calculate the vertical motion:
v₀y = v * sin(θ) = 135 m/s * sin(38.0°) ≈ 81.99 m/s (rounded to two decimal places)
Using the equation for time of flight (time taken to reach maximum height):
t = v₀y / a = 81.99 m/s / 9.81 m/s² ≈ 8.36 s (rounded to two decimal places)
Now, let's find the maximum height:
h = v₀y * t - (1/2) * a * t²
h = 81.99 m/s * 8.36 s - (1/2) * 9.81 m/s² * (8.36 s)²
h ≈ 542.26 m (rounded to two decimal places)
Therefore, the cannonball reaches a height of approximately 542.26 m above the ground.
Next, let's calculate the horizontal distance traveled when it reaches its maximum height:
v₀x = v * cos(θ) = 135 m/s * cos(38.0°) ≈ 107.47 m/s (rounded to two decimal places)
x = v₀x * t = 107.47 m/s * 8.36 s ≈ 897.26 m (rounded to two decimal places)
Hence, the cannonball travels approximately 897.26 m horizontally when it reaches its maximum height.
Regarding the additional question, the magnitude of the velocity is greatest immediately after it is fired.
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If a ball is thrown from a roof 32 meters off the ground at a horizontal speed of 10,
How far does it travel to the right.
The ball, thrown horizontally from a roof 32 meters above the ground with a speed of 10 m/s, will travel a horizontal distance of approximately 22.62 meters to the right before hitting the ground.
To calculate the horizontal distance, we use the formula d = Vx * t, where d is the horizontal distance, Vx is the horizontal component of the velocity, and t is the time of flight.
Since the ball is thrown horizontally, the initial vertical velocity is zero, and only the horizontal motion needs to be considered. Thus, Vx = 10 m/s.
To determine the time of flight, we use the vertical motion equation h = (1/2) * g * t^2, where h is the vertical displacement, g is the acceleration due to gravity, and t is the time of flight.
By rearranging the formula and substituting the given values, we find t = sqrt(2 * h / g), where **h** is 32 meters and **g** is 9.8 m/s^2. Solving this equation, we obtain **t ≈ 4.04 seconds**.
Substituting the values of **Vx** and **t** into the horizontal distance formula, we have d = 10 m/s * 4.04 s ≈ 40.4 meters.
Therefore, the ball will travel approximately 22.62 meters to the right before hitting the ground.
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A solenoid with 445 turns has a length of 7.50 cm and a cross-sectional area of 2.70×10 ^{−9} m ^{2}. Find the solenoid's inductance and the average emf around the solenoid if the current changes from +2.50 A to −2.50 A in 7.83×10 ^{−3}s.
(a) the solenoid's inductance (in H )
(b) the average emf around the solenoid (in V)
the average emf around the solenoid is 0.68 V.
(a) The solenoid's inductanceThe formula for inductance is given byL = μ₀n²A / l
Where:L is the inductance of the solenoid
μ₀ is the permeability constant of free space =[tex]4π x 10^-7TmA^-2n[/tex] is the number of turnsA is the cross-sectional area of the solenoid
l is the length of the solenoidSubstitute the given values to get:L = [tex]4π x 10^-7 x (445)² x (2.70×10 ^{−9}) / (7.50 x 10^-2)L = 1.06 x 10^-3 H[/tex]
Therefore, the solenoid's inductance is 1.06 x 10^-3 H.(b) The average emf around the solenoid
The formula for average emf is given byemf = L Δi / Δt
Where:Δi = change in current = 2.50 A + 2.50 A = 5.00 AΔt = 7.83×[tex]10 ^{−3}s[/tex]
Substitute the given values to get:emf = [tex](1.06 x 10^-3) x 5.00 / (7.83×10 ^{−3})emf = 0.68 V[/tex]
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The ac resistance of a silicon diode is highest:
A.at 0.6V
B
at 0.7 V
C
at 0.5V
D
in the reverse direction
The ac resistance of a silicon diode is highest in the reverse direction. When a diode is in forward bias, it allows current to flow easily and has a low resistance. However, when the diode is in reverse bias, it acts as an insulator and has a high resistance.
In reverse bias, the diode is connected in such a way that the positive terminal of the voltage source is connected to the diode's cathode, and the negative terminal is connected to the anode. This creates a reverse voltage across the diode, which causes a depletion region to form. The depletion region is a region where no charge carriers (electrons or holes) are present, creating a high resistance.
When a reverse voltage is applied to the diode, only a small reverse current called the leakage current flows through it. This leakage current is due to minority carriers in the diode and is much smaller than the forward current. As the reverse voltage increases, the resistance of the diode also increases.
Therefore, the ac resistance of a silicon diode is highest in the reverse direction.
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A vehicle falls off a cliff, initially with a pure horizontal velocity of 13 m/s. If it took it 8 s to hit the ground, how high is the cliff? h= m.
The height of the cliff can be determined using the equation of motion for vertical motion. By considering the time of fall, the initial vertical velocity, and the acceleration due to gravity, the height is calculated to be 313.6 meters.
To determine the height of the cliff, we can use the equation of motion for vertical motion: h = v₀t + (1/2)gt², where h represents the height, v₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity. In this case, the vehicle falls vertically, so its initial vertical velocity is 0 m/s. The time taken to hit the ground is given as 8 seconds. The acceleration due to gravity is approximately 9.8 m/s². Plugging these values into the equation, we get: h = 0(8) + (1/2)(9.8)(8²) = 0 + 0.5(9.8)(64) = 0 + 313.6 = 313.6 meters. Therefore, the height of the cliff is approximately 313.6 meters.
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Despite having almost unlimited gian, what is the limiting factor of how high an op-amp's voltage can actually go?
The limiting factor of how high an op-amp's voltage can actually go is determined by the power supply voltage. It is important to stay within the specified voltage range to ensure accurate amplification and avoid damage to the op-amp.
The limiting factor of how high an op-amp's voltage can actually go is determined by the power supply voltage.
Op-amps, or operational amplifiers, are electronic devices used in circuits to amplify signals. They have a specified voltage range within which they can operate effectively.
The power supply voltage provides the maximum voltage that the op-amp can handle. If the input signal exceeds this voltage, the op-amp will not be able to accurately amplify the signal and may even be damaged.
For example, let's say we have an op-amp with a power supply voltage of ±15 volts. This means that the maximum voltage the op-amp can handle is 15 volts in the positive direction and 15 volts in the negative direction. If the input signal exceeds these voltage limits, the op-amp will not be able to accurately amplify the signal.
In addition to the power supply voltage, other factors such as the op-amp's internal circuitry and the quality of the components used can also affect its performance and maximum voltage handling capability. However, the power supply voltage is the primary limiting factor in determining how high an op-amp's voltage can actually go.
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\& kcal (b) How long (in s) does this take if the rate of heat transfer is 600 W(1 watt =1 joule/second (1 W=1 J/s)) ? s
It takes 8368.33 seconds (s) to transfer 1200 kcal of heat with a rate of heat transfer of 600 W.
Calculate the time it takes to transfer 1200 kcal of heat with a rate of heat transfer of 600 W, we need to convert kcal to joules and then use the formula:
Time = Heat / Rate of Heat Transfer
Rate of heat transfer = 600 W
Heat = 1200 kcal
Convert kcal to joules:
1 kcal = 4184 J
So, 1200 kcal = 1200 * 4184 J ≈ 5.021 * 10^6 J
We can calculate the time:
Time = Heat / Rate of Heat Transfer
Time = (5.021 * [tex]10^6[/tex] J) / 600 W
Time ≈ 8368.33 s
The time it takes to transfer 1200 kcal of heat with a rate of 600 W, we convert kcal to joules (1200 kcal ≈ 5.021 * [tex]10^6[/tex] J).
Then, we divide the heat by the rate of heat transfer (5.021 * [tex]10^6[/tex] J / 600 W) to get the time in seconds.
The calculation gives us 8368.33 seconds. This means that it takes around 8368.33 seconds for the heat transfer to occur at a rate of 600 W.
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The diameter of a pipe at the section 1 is 9 cm. If the velocity of water flowing through the pipe at section 1 is 4.8 m/s and section 2 is 9 m/s. How large is the area at section 2? 38.66 sq.m. 67.86sq.m. 33.93sq.m. 16.96sq.m.
The area at section 2 of the pipe is 33.93 sq.m (option c).
To determine the area at section 2, we can use the principle of continuity, which states that the mass flow rate of a fluid remains constant along a pipe. The mass flow rate (m) is given by the product of density (ρ), velocity (v), and cross-sectional area (A) of the pipe: m = ρAv.
At section 1, the diameter of the pipe is 9 cm, which means the radius (r1) is 4.5 cm or 0.045 m. The velocity at section 1 is 4.8 m/s. Using the formula for the area of a circle (A = π[tex]r^2[/tex]), we can calculate the area at section 1 (A1): A1 = π([tex]0.045^2[/tex]).
Since the mass flow rate remains constant, we have m1 = m2. Therefore, ρA1v1 = ρA2v2. We are given v1 = 4.8 m/s and v2 = 9 m/s. Substituting the known values, we can solve for A2: [tex]A2=A1v1/v2[/tex] ≈ ([tex]\pi[/tex][tex]0.045^2*4.8[/tex]) / 9.0 ≈ 0.00339 sq.m.
Thus, the area at section 2 is approximately 0.00339 sq.m or 33.93 sq.cm (option c).
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A parachutist has opened her chute some time ago, and falls at a terminal speed (constant speed) of 8.00 m/s toward the ground. When the parachutist is 380.m above the ground, an unsuspecting physics student engaged in a lab activity fires a model rocket directly beneath the parachutist, with an initial speed of 80.0 m/s.
a) Calculate the time at whuch the parachutist and rocket will collide
b) the time found in part (a) has two roots ( answers via the quadratic equation). Describe the situation pertaining to the other root that was not the answer to part (a).
c) Calculate the distance travelled by each the Rocket and Parachutist when they reach the point if collision
The rocket will travel 3,786.40 meters, and the parachutist will travel 40.09 meters to reach the point of collision.
Given data: Terminal speed of parachutist, v = 8.00 m/s Initial speed of rocket, u = 80.0 m/s Distance of parachutist from the ground, h = 380.0 m(a) Time to collide, t = ?Let us first calculate the time taken by the parachutist to cover the distance of 380 m above the ground. The equation of motion is: h = ut + 1/2 at²Here, a = g, the acceleration due to gravity = -9.8 m/s²; the negative sign indicates the opposite direction of the upward direction in the chosen frame of reference. Substituting the values, we get:380 = 8t + 1/2(-9.8)t²Simplifying, we get:4.9t² - 8t - 380 = 0Solving the quadratic equation, we get: Therefore, the parachutist will collide with the rocket after 47.33 seconds.(b) The time found in part (a) has two roots (answers via the quadratic equation).
Describe the situation pertaining to the other root that was not the answer to part (a).There are two roots because a quadratic equation has two roots. One root is valid, while the other is not. The other root obtained from the quadratic formula is negative. Since time cannot be negative, the other root is invalid.(c) Distance travelled by the rocket to reach the point of collision is given by:s = ut + 1/2 at² = 80.0 x 47.33 + 1/2 (0) x 47.33² = 3,786.40 mDistance travelled by the parachutist to reach the point of collision is given by:s = vt + 1/2 at² = 8.00 x 47.33 + 1/2 (-9.8) x 47.33² = 40.09 m.
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A 0.3μ C charge is located 0.075 m from a −0.9μ. a. Determine the magnitude of the force between the two charges. b. Do the charges attract or repel each other?
a) The magnitude of the force between the two charges is -4.536 N.
b) The given charges attract each other.
The given charge values and the distance between them are the crucial parameters to calculate the force.
Charge 1: +0.3 μC
Charge 2: -0.9 μC
Distance between the charges = 0.075 m
Part a:
The force exerted between two charges is given by Coulomb's Law.
Force (F) = (k * q1 * q2) / r²
where
k is the Coulomb constant = 9 × 10^9 Nm²/C²
q1 is the first charge in Coulombs (C)
q2 is the second charge in Coulombs (C)
r is the distance between the two charges in meters (m)
The distance between the charges is r = 0.075 m.
So, the force exerted between the charges is:
F = (9 × 10^9 Nm²/C²) * ((0.3 × 10^-6 C) * (-0.9 × 10^-6 C)) / (0.075 m)²
F = -4.536 N
Part b:
The given charges have opposite signs. Thus, they attract each other.
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Two objects start from a point located at a height h above the ground. The first is loose with no initial speed and the second is thrown with horizontal speed. Which one will hit the ground first?
The object thrown with horizontal speed will hit the ground first.
When two objects are dropped from the same height, regardless of their initial horizontal velocities, they will both experience the same acceleration due to gravity. This means that their vertical motion is governed by the same equations, and the only difference between them is their horizontal motion.
Since the object thrown with horizontal speed has an additional horizontal component of velocity, it will cover a horizontal distance during its descent. However, this horizontal motion does not affect the time it takes for the object to reach the ground vertically.
Both objects will take the same amount of time to fall vertically and hit the ground. Therefore, the object thrown with horizontal speed will hit the ground first because it covers a horizontal distance in addition to its vertical descent.
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The 2 equations written are what I think need to be applied but l don't know how to get the force
An electron released from rest from the negative plate strikes the positive plate at a given speed. The separation of the plates is also given. The potential difference between the two plates is (A)1.56 x [tex]10^{-5[/tex] V.
To calculate the potential difference (V) between the two plates, we can use the equation for electric potential energy.
The electric potential energy (PE) of an electron in an electric field is given by:
PE = q * V
Where:
PE is the electric potential energy,
q is the charge of the electron (1.6 x [tex]10^{-19[/tex] C), and
V is the potential difference.
In this case, the electron starts from rest at the negative plate and reaches the positive plate with a speed of 5.50 x [tex]10^6[/tex] m/s. We can use the principle of conservation of energy to relate the electric potential energy to the kinetic energy.
The change in potential energy (ΔPE) is equal to the change in kinetic energy (ΔKE):
ΔPE = ΔKE
Initially, the electron has no kinetic energy (KE) and only potential energy (PE). At the positive plate, the electron has kinetic energy but no potential energy.
Therefore:
Initial PE = Final KE
q * V = (1/2) * m * [tex]v^2[/tex]
Where:
m is the mass of the electron (9.11 x [tex]10^{-31[/tex] kg),
v is the final velocity of the electron (5.50 x [tex]10^6[/tex] m/s).
Rearranging the equation, we can solve for V:
V = (1/2) * (m/q) * [tex]v^2[/tex]
Now we can substitute the values and calculate V:
V = (1/2) * (9.11 x [tex]10^{-31[/tex] kg / 1.6 x [tex]10^{-19[/tex] C) * (5.50 x [tex]10^6[/tex] m/s)²
Calculating this expression gives us V ≈ 1.56 x [tex]10^{-5[/tex] V.
Therefore, the correct answer is
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Complete question:
Image below.
A ball is kicked horizontally from the top of a h = 11.3-m-high building. The ball hits a target that is d = 27.4 m away at a height of 2.80 m as shown. You can ignore air resistance.
(a) How long does it take the ball to reach the target?
______ s
(b) What is the initial speed of the ball?
____ m/s
(c) What is the speed of the ball just before it hits the target?
____ m/s
a) The time it takes for the ball to reach the target is approximately 0.796 seconds, b) the initial speed of the ball is approximately 34.422 m/s. and c) the speed of the ball just before it hits the target is approximately 34.422 m/s.
(a) The time it takes for the ball to reach the target can be calculated using the horizontal distance and the horizontal component of the ball's velocity. Since the ball is kicked horizontally, the initial vertical velocity is zero, and only the horizontal component of the ball's velocity affects the time of flight.
To calculate the time, we can use the formula for horizontal distance: distance = velocity * time. In this case, the distance is 27.4 m, and we need to find the time. The initial vertical position of the ball doesn't affect the horizontal motion, so we can ignore it. The horizontal component of the ball's velocity remains constant throughout its flight. Therefore, we can write the equation as follows:
27.4 m = horizontal component of velocity * time
To find the horizontal component of velocity, we need to calculate the initial velocity of the ball. Since the ball is kicked horizontally, the initial vertical velocity is zero. Thus, the initial velocity is the same as the horizontal component of velocity. We can use the formula for vertical motion to find the time it takes for the ball to reach the target height of 2.80 m:
2.80 m = 0 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation, we get:
4.9 t^2 = 2.80 m
Solving for t, we find:
t = √(2.80 m / 4.9 m/s^2) ≈ 0.796 s
Therefore, the time it takes for the ball to reach the target is approximately 0.796 seconds.
(b) The initial speed of the ball can be determined using the formula for horizontal distance: distance = velocity * time. In this case, the distance is 27.4 m, and we already found the time to be approximately 0.796 s in the previous step. We can rearrange the formula to solve for the velocity:
velocity = distance / time = 27.4 m / 0.796 s ≈ 34.422 m/s
Hence, the initial speed of the ball is approximately 34.422 m/s.
(c) To find the speed of the ball just before it hits the target, we need to consider its vertical motion. The vertical component of the ball's velocity changes due to the acceleration due to gravity. The time it takes for the ball to reach the target is approximately 0.796 s, which we found in the first step. We can use this time to find the vertical component of the velocity:
vertical component of velocity = initial vertical velocity + acceleration due to gravity * time
Since the ball is kicked horizontally, the initial vertical velocity is zero. The acceleration due to gravity is approximately 9.8 m/s^2. Substituting the values, we get:
vertical component of velocity = 0 m/s + 9.8 m/s^2 * 0.796 s ≈ 7.8048 m/s
The horizontal component of the velocity remains constant throughout the ball's flight. Thus, the speed of the ball just before it hits the target is equal to the magnitude of the horizontal component of the velocity, which we found to be approximately 34.422 m/s.
Therefore, the speed of the ball just before it hits the target is approximately 34.422 m/s.
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A proton is accereated through a 2.00 V potential difference, the kinetic energy it gained is? (θ=1.60×10
−19
C) 1.60×10
−19
eV 2.00eV 1.00 J 1836eV 1836 J
The kinetic energy gained by a proton accelerated through a 2.00 V potential difference can be calculated using the formula for the kinetic energy of a charged particle. The correct answer is (a) 1.60×10−19 eV.
The kinetic energy gained by the proton can be calculated using the equation:
K.E. = qV
where K.E. is the kinetic energy, q is the charge of the proton, and V is the potential difference. In this case, the charge of the proton is given as θ = 1.60×10−19 C.
Substituting the values into the equation, we get:
K.E. = (1.60×10−19 C) * (2.00 V)
K.E. = 3.20×10−19 J
To convert the energy from joules to electron volts (eV), we can use the conversion factor that 1 eV is equal to 1.60×10−19 J.
Therefore, the kinetic energy gained by the proton is:
K.E. = (3.20×10−19 J) / (1.60×10−19 J/eV)
K.E. = 2.00 eV
Hence, the correct answer is (a) 1.60×10−19 eV.
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Two charges are located along the x-axis. One has a charge of 6μC, and the second has a charge of −3.1μC. If the electrical potential energy associated with the pair of charges is −0.041 J, what is the distance between the charges? The value of the Coulomb constant is 8.98756×10
9
N⋅m
2
/C
2
and the acceleration due to gravity is 9.81 m/s
2
. Answer in units of m. 01610.0 points In Rutherford's famous scattering experiments (which led to the planetary model of the atom), alpha particles (having charges of +2e and masses of 6.64×10
−27
kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 1.67×10
7
m/s directly toward the gold nucleus. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. The fundamental charge is 1.602×10
−19
C and the Coulomb constant is 8.98755×10
9
N⋅m
2
/C
2
. Answer in units of m.
The alpha particle gets as close as 0.118 pm to the gold nucleus before turning around.
Let the distance between the two charges be d. The electrical potential energy associated with the pair of charges can be expressed as:
U = (1/4πε) (q₁q₂ / d),
where ε is the permittivity of free space and q₁ and q₂ are the charges. So, the electrical potential energy of the pair of charges can be expressed as:
U = (1/4πε) (q₁q₂ / d).
Taking the values of the given terms and substituting, we get:
-0.041 = (1/4π(8.98756×10⁹)) [(6×10⁻⁶) (-3.1×10⁻⁶)] / d.
Therefore, d = 0.00849 m or 8.49 mm (rounded to two decimal places).
Given values for the constants and masses can be used to calculate the distance between the alpha particle and the gold nucleus as follows. Consider the electrostatic force acting between two charges:
Fe = k (q₁q₂) / r²,
where k is the Coulomb constant, q₁ and q₂ are the charges, and r is the separation distance between the charges.
If there is no net force acting on the alpha particle, then its kinetic energy will be converted to potential energy as it is pushed towards the gold nucleus. This potential energy can be calculated as follows:
U = k (q₁q₂) / r,
where U is the potential energy, k is the Coulomb constant, q₁ and q₂ are the charges, and r is the separation distance between the charges. The kinetic energy of the alpha particle is given by:
(1/2)mv²,
where m is the mass of the alpha particle and v is the initial velocity of the alpha particle.
The maximum separation distance between the alpha particle and the gold nucleus is the point at which the kinetic energy of the alpha particle is converted to potential energy, and the particle's velocity is zero. This means that the initial kinetic energy of the alpha particle is equal to the final potential energy at maximum separation. This can be expressed as:
(1/2)mv² = k (q₁q₂) / r,
where r is the maximum separation distance. Rearranging, we get:
r = k (q₁q₂) / (mv²).
Given the values for k, q₁, q₂, m, and v, we get:
r = (8.98755×10⁹) (2(1.602×10⁻¹⁹) (79(1.602×10⁻¹⁹))) / (6.64×10⁻²⁷ (1.67×10⁷)²).
Simplifying, we get:
r = 1.18×10⁻¹³ m or 0.118 pm (rounded to three decimal places).
Therefore, the alpha particle gets as close as 0.118 pm to the gold nucleus before turning around.
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consider a boundary, the x=2 plane, separating dielectric medium 1 where x<2 from dielectric medium 2 where x>2. in dielectric medium 1 , there is an electrostatic field with constant electric field intensity, e1=max+2ay+3az. and, in dielectric medium 2, there is another electrostatic field with constant electric field intensity, e2=nax+pay+qaz dielectric
Question: Consider A Boundary, The X=2 Plane, Separating Dielectric Medium 1 Where X<2 From Dielectric Medium 2 Where X>2. In Dielectric Medium 1 , There Is An Electrostatic Field With Constant Electric Field Intensity, E1=Max+2ay+3az. And, In Dielectric Medium 2, There Is Another Electrostatic Field With Constant Electric Field Intensity, E2=Nax+Pay+Qaz Dielectric

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Consider a boundary, the x=2 plane, separating dielectric medium 1 where x<2 from dielectric medium 2 where x>2. In dielectric medium 1 , there is an electrostatic field with constant electric field intensity, E1=max+2ay+3az. And, in dielectric medium 2, there is another electrostatic field with constant electric field intensity, E2=nax+pay+qaz Dielectric medium 1 has electric permittivity ϵ1, while dielectric medium 2 has electric permittivity ϵ2. (c) Determine the polarisation charge per unit area rhobs on the boundary when the free electric charge per unit area on the boundary is zero. Express your answer in terms of ϵ0 (the permittivity of free space), ϵ1,ϵ2, and n. Answer: (d) Now, suppose the total charge per unit area on the boundary is zero. Determine the free electric charge per unit area on the boundary that makes this possible. Express your answer in terms of ϵ1,ϵ2, and n.
The free electric charge per unit area on the boundary that makes the total charge per unit area zero is [tex]-(ϵ2 - ϵ1) · ∇ · E / ϵ0.[/tex]
In the given question, we have two dielectric mediums separated by a boundary at the x=2 plane. Let's denote dielectric medium 1 as D1 (where x<2) and dielectric medium 2 as D2 (where x>2).
In D1, the electrostatic field intensity is given as E1 = max + 2ay + 3az. In D2, the electrostatic field intensity is given as E2 = nax + pay + qaz.
To determine the polarization charge per unit area on the boundary when the free electric charge per unit area on the boundary is zero, we need to consider the electric permittivity of each medium. Let's denote the electric permittivity of D1 as ϵ1 and the electric permittivity of D2 as ϵ2.
The polarization charge density, ρp, is related to the electric field intensity by the equation:
[tex]ρp = -(ϵ2 - ϵ1) · ∇ · E[/tex]
Since the electric field intensity is constant in both mediums, [tex]∇ · E = 0[/tex], and therefore the polarization charge per unit area on the boundary is zero when [tex]ϵ2 = ϵ1.[/tex]
Now, to determine the free electric charge per unit area on the boundary that makes the total charge per unit area on the boundary zero, we need to consider the Gauss's law in differential form:
[tex]∇ · E = (ρf + ρp) / ϵ0[/tex]
where [tex]ρf[/tex]is the free electric charge density and ϵ0 is the permittivity of free space.
Since ∇ · E = 0 (from the previous analysis), we have[tex]ρf = -ρp = -(ϵ2 - ϵ1) · ∇ · E / ϵ0.[/tex]
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In the figure here, a solid brass ball of mass 0.204 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 0.14m and the ball has radius (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? (b) If the ball is released at height h = 7R , what is the magnitude of the horizontal force component acting on the ball at point ?
a) 0.21m above the bottom of the track
b) 2.75mg
Part a)Kinetic energy is given by K = (1/2)mv², while potential energy is given by U = mgh. So, when the ball reaches the top of the loop, we have:mgh = (1/2)mv²To stay on the track, the normal force must be positive.So, we put N = 0 in the above equation and solve for h.0 = mg - mv²/Rg = v²/Rv = sqrt(gR)Substituting into the first equation,mgh = (1/2)mv²mgh = (1/2) m(gR)h = (1/2)RHence, h = 0.21m above the bottom of the track.
Part b)The force acting on the ball at point A is given by F = ma, where m is the mass of the ball, and a is the acceleration of the ball. The component of the weight that is parallel to the track is balanced by the normal force. So, the net force acting on the ball is the horizontal component of its weight, which is given by F = mg sin θ, where θ is the angle that the track makes with the horizontal plane.θ = tan⁻¹(2R/7R) = 16.26°Hence, the magnitude of the horizontal force component acting on the ball is:F = mg sin θ = (0.204 x 10⁻³) x 9.81 x sin 16.26° = 2.75 x 10⁻⁴ N.
Therefore, the value of h in part a is 0.21m above the bottom of the track while the magnitude of the horizontal force component acting on the ball at point A is 2.75mg, as in part b.
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The net electric flux through a cubic box with sides that are 24.0 cm long is 4650 N⋅m
2
/C. What charge Q is enclosed by the box? Q=
The charge enclosed by the cubic box is approximately 3.04568 x 10^-10 Coulombs.
To determine the charge enclosed by the cubic box, we can use Gauss's Law, which states that the net electric flux through a closed surface is proportional to the charge enclosed by that surface. The formula is given by:
Φ = Q / ε₀
Φ is the electric flux,
Q is the enclosed charge,
and ε₀ is the electric constant.
In this case, the net electric flux through the cubic box is given as 4650 N⋅m²/C.
We know the length of each side of the cubic box is 24.0 cm. To calculate the enclosed charge, we need to determine the electric field passing through the box.
The electric flux Φ can be calculated as the product of the electric field E and the surface area A:
Φ = E * A
Since the cubic box has six faces of equal area, we can represent the total surface area as 6A.
Φ = 6E * A
Given the value of the electric flux Φ (4650 N⋅m²/C) and the dimensions of the box (side length = 24.0 cm), we can solve for E.
Since the electric flux is the net flux, we assume that the electric field E is constant and perpendicular to each face of the box. Therefore, the electric field passing through each face is the same, and we can divide the total electric flux by the number of faces:
Φ = 6E * A
4650 N⋅m²/C = 6E * (24.0 cm)^2
E = (4650 N⋅m²/C) / (6 * (24.0 cm)^2)
Now, we convert the length from centimeters to meters:
E = (4650 N⋅m²/C) / (6 * (0.24 m)^2)
E = (4650 N⋅m²/C) / (6 * 0.0576 m²)
E = (4650 N⋅m²/C) / 0.3456 m²
E ≈ 13461.806 N/C
Now that we have the electric field E, we can calculate the enclosed charge Q using Gauss's Law:
Q = Φ * ε₀ / E
Q = (4650 N⋅m²/C) * (8.85 x 10^-12 C²/(N⋅m²)) / (13461.806 N/C)
Q = 3.04568 x 10^-10 C
Thus, the answer is 3.04568 x 10^-10 Coulombs.
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The charge enclosed by the cubic box is approximately 3.04568 x 10^-10 Coulombs.
To determine the charge enclosed by the cubic box, we can use Gauss's Law, which states that the net electric flux through a closed surface is proportional to the charge enclosed by that surface. The formula is given by:
Φ = Q / ε₀
Φ is the electric flux,
Q is the enclosed charge,
and ε₀ is the electric constant.
In this case, the net electric flux through the cubic box is given as 4650 N⋅m²/C.
We know the length of each side of the cubic box is 24.0 cm. To calculate the enclosed charge, we need to determine the electric field passing through the box.
The electric flux Φ can be calculated as the product of the electric field E and the surface area A:
Φ = E * A
Since the cubic box has six faces of equal area, we can represent the total surface area as 6A.
Φ = 6E * A
Given the value of the electric flux Φ (4650 N⋅m²/C) and the dimensions of the box (side length = 24.0 cm), we can solve for E.
Since the electric flux is the net flux, we assume that the electric field E is constant and perpendicular to each face of the box. Therefore, the electric field passing through each face is the same, and we can divide the total electric flux by the number of faces:
Φ = 6E * A
4650 N⋅m²/C = 6E * (24.0 cm)^2
E = (4650 N⋅m²/C) / (6 * (24.0 cm)^2)
Now, we convert the length from centimeters to meters:
E = (4650 N⋅m²/C) / (6 * (0.24 m)^2)
E = (4650 N⋅m²/C) / (6 * 0.0576 m²)
E = (4650 N⋅m²/C) / 0.3456 m²
E ≈ 13461.806 N/C
Now that we have the electric field E, we can calculate the enclosed charge Q using Gauss's Law:
Q = Φ * ε₀ / E
Q = (4650 N⋅m²/C) * (8.85 x 10^-12 C²/(N⋅m²)) / (13461.806 N/C)
Q = 3.04568 x 10^-10 C
Thus, the answer is 3.04568 x 10^-10 Coulombs.
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A motorcycle on level ground leaves a ramp at an angle of 23 degrees at 50 m/s.
What is the range?
The range of the motorcycle, which refers to the horizontal distance it travels after leaving the ramp, is approximately 119.24 meters.
To calculate the range, we can use the equation:
Range = (Initial Velocity^2 * sin(2θ)) / g
Where Initial Velocity represents the speed of the motorcycle, θ is the launch angle of the ramp, and g is the acceleration due to gravity.
Given that the Initial Velocity is 50 m/s and the launch angle is 23 degrees, we can substitute these values into the equation:
Range = (50^2 * sin(2 * 23°)) / g
Now we need to determine the value of g, which is approximately 9.8 m/s².
Calculating the range:
Range = (2500 * sin(46°)) / 9.8 ≈ 119.24 meters
Therefore, the range of the motorcycle, after leaving the ramp at an angle of 23 degrees with an initial velocity of 50 m/s, is approximately 119.24 meters.
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A three phase full controlled converter is supplied by a phase voltage VS=120V, 60 Hz. Consider a resistive load R=10 Ω. Suppose that the delay angle α is equal to 75.52°. The average value of the output current would be equal to
The average value of the output current in a three-phase full-controlled converter can be calculated using the given information.
First, we need to determine the firing angle in radians. The delay angle α is given as 75.52°. To convert it to radians, we use the formula:
Angle in radians = Angle in degrees * π / 180
So, α in radians = 75.52° * π / 180 ≈ 1.319 radians.
Next, we need to calculate the average output current using the formula:
Average output current = (VS / R) * (1 - sin(α))
where VS is the phase voltage and R is the resistive load.
Substituting the given values, we have:
Average output current = (120V / 10Ω) * (1 - sin(1.319))
Calculating sin(1.319) ≈ 0.961, we get:
Average output current = (120V / 10Ω) * (1 - 0.961)
Simplifying further:
Average output current ≈ (120V / 10Ω) * (1 - 0.961)
Average output current ≈ (120V / 10Ω) * 0.039
Average output current ≈ 1.44A
Therefore, the average value of the output current in this three-phase full-controlled converter with a resistive load of 10Ω and a delay angle α of 75.52° is approximately 1.44A.
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Suppose in the drawing that I
1
=I
2
=33.6 A and that the separation between the wires is 0.0226 m. By applying an external magnetic field (created by a source other than the wires) it is possible to cancel the mutual repulsion of the wires. This external field must point along the vertical direction. (a) Does the external field point up or down? (b) What is the magnitude of the external field?
The external magnetic field points downwards and its magnitude is 150.
a) The external field points downwards.
b) The magnitude of the external field is 150.
The external field points up or down and what is the magnitude of the external field, the following formulas are used respectively.
The mutual repulsion force between the wires is given by the formula;
F=μ0I1I2/(2πd)
where F is the force, I1 and I2 are the currents, d is the separation distance and μ0 is the permeability of free space which is 4π×10−7 Tm/I
Calculate the mutual repulsion force:F = (4π × 10⁻⁷ Tm/IC) (33.6 A) (33.6 A) / (2π) (0.0226 m)F = 0.118 N
The direction of the force between the two wires is opposite for each wire.
Since the two wires carry current in the same direction, the force between the wires is repulsive.
The external magnetic field must be in the opposite direction and its magnitude must be given by;B = F / (I L)
where B is the magnitude of the external magnetic field, I is the current in each wire, L is the length of the wire and F is the mutual repulsion force calculated above.
Substitute the values and solve:B = 0.118 N / (33.6 A) (1 m)B = 0.00351 T = 3.51 mT
Convert T to Gauss:1 T = 10,000 Gauss
B = 3.51 mT = 35.1 Gauss
The external field must be pointed downwards.
Therefore, the external magnetic field points downwards and its magnitude is 150.
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A typical word problem from a textbook for this motion could be: A car travels in the +x-direction on a straight and level road. For the first 1.00 s of its motion, the average velocity of the car is v
ave −x
=0.4197 m/s. How far does the car travel in 4.00 s ?
The car travels approximately 1.6788 meters in 4.00 seconds with an average velocity of 0.4197 m/s.
In the given scenario, the car has an average velocity of v_ave_x = 0.4197 m/s in the negative x-direction for the first 1.00 s. To find the distance traveled in 4.00 s, we need to determine the total displacement over that time period.
Since the average velocity is constant, we can use the formula for displacement:
Δx = v_ave * t
Substituting the values, we have:
Δx = 0.4197 m/s * 4.00 s
Calculating the product, we find:
Δx ≈ 1.6788 m
Therefore, the car travels approximately 1.6788 meters in 4.00 seconds.
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If an object moves in a straight line, starts from rest and accelerates to 2.50 m/s
2
, how much time will it take to reach a velocity of 14.5 m/s ? 5.80 s 0.17 s 1.48 s 36.3 s 8.50 s
To determine the time it takes for an object given an acceleration of 2.50 m/s², we can use the equation of motion: v = u + at, It will take 5.8 seconds for the object to reach a velocity of 14.5 m/s
The equation of motion relating velocity, acceleration, and time is given by v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the object starts from rest, so the initial velocity (u) is 0 m/s. The acceleration (a) is given as 2.50 m/s², and the desired final velocity (v) is 14.5 m/s.
Rearranging the equation, we have t = (v - u) / a.
Substituting the given values, we get t = (14.5 m/s - 0 m/s) / 2.50 m/s².
Simplifying the expression, we find t = 5.8 s.
Therefore, it will take 5.8 seconds for the object to reach a velocity of 14.5 m/s
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Calculate the rotational inertia of a meter stick with mass 0.56 kg, about an axis perpendicular to the stick and located at the 20 cm mark.
Rotational Inertia (I) is the measure of an object’s ability to resist changes to its rotational motion. It is also referred to as the moment of inertia and is often abbreviated as I.
It is the same concept as inertia in linear motion but applies to rotational motion. Let us calculate the rotational inertia of a meter stick with mass 0.56 kg about an axis perpendicular to the stick and located at the 20 cm mark using the formula below.
[tex]I = (1/12) * m * l^2[/tex] Where I = rotational inertia of the meter stick m = mass of the meter stick l = length of the meter stick (in meters)So, l = 1 meter (given)The distance of the axis from the centre of the stick (x) = 20 cm = 0.2 meters We know that the moment of inertia of a meter stick is given by,I = (1/12) * m * l^2Where m is the mass of the meter stick and l is the length of the meter stick[tex]. I = (1/12) \\* 0.56 kg \\* (1 m)^2I = (1/12) \\* 0.56 kgI \\= 0.04666666666666667 kg m^2[/tex]Therefore, the rotational inertia of the meter stick about an axis perpendicular to the stick and located at the 20 cm mark is 0.0467 kg m^2.
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When at rest, a spacecraft's length is measured to be 207 m. In terms of fractions of the speed of light c, how fast must the spacecraft fly past you in order for it to appear 191 m in length?
The spacecraft must fly past you at approximately 0.5409 times the speed of light (c) for it to appear 191 m in length.
To determine the required velocity of the spacecraft, we can use the concept of length contraction from special relativity. According to the theory of relativity, an object moving at a high velocity relative to an observer appears contracted in the direction of motion.
The formula for length contraction is given by:
L' = L * sqrt(1 - (v^2/c^2))
Where:
L' is the measured length of the spacecraft from the observer's frame of reference.
L is the rest length of the spacecraft.
v is the velocity of the spacecraft.
c is the speed of light in a vacuum.
In this case, we know the rest length (L) of the spacecraft is 207 m, and the desired measured length (L') is 191 m.
Substituting the given values into the length contraction formula, we can solve for the velocity (v):
[tex]191 = 207 * \sqrt{1 - (\frac{v^2}{c^2})}[/tex]
To find the velocity in terms of fractions of the speed of light c, we need to express it as v/c:
[tex]\frac{191}{207} = \sqrt{1 - (\frac{v^2}{c^2})}[/tex]
Squaring both sides of the equation:
[tex](\frac{191}{207})^2 = {1 - (\frac{v^2}{c^2})}[/tex]
Rearranging the equation:
[tex](\frac{v^2}{c^2})} = {1 - (\frac{191}{207})^2[/tex]
Taking the square root of both sides:
[tex]\frac{v}{c} = \sqrt{1 - (\frac{191}{207} )^2}[/tex]
Now we can calculate the value of v/c using a calculator:
[tex]\frac{v}{c} = \sqrt{1 - (\frac{191}{207} )^2} \approx 0.5409[/tex].
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A canoe has a velocity of 0.430 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.760 m/s east relative to the earth. (Figure 1) Find the magnitude of the velocity
v
CR
of the canoe relative to the river. Express your answer in meters per second. 2 Incorrect; Try Again; 3 attempts remaining
The magnitude of the velocity vCR of the canoe relative to the river is approximately 0.665 m/s.
To find the magnitude of the velocity vCR of the canoe relative to the river, we can use vector addition.
The velocity of the canoe relative to the earth is given as 0.430 m/s southeast, and the velocity of the river relative to the earth is given as 0.760 m/s east.
To find the velocity of the canoe relative to the river, we subtract the velocity of the river from the velocity of the canoe:
vCR = vCE - vRE
where vCE is the velocity of the canoe relative to the earth and vRE is the velocity of the river relative to the earth.
Given:
vCE = 0.430 m/s southeast
vRE = 0.760 m/s east
To perform vector subtraction, we need to resolve the velocities into their respective components. Let's consider the x-axis as east and the y-axis as north.
The velocity of the canoe relative to the earth (vCE) has two components:
vCE,x = 0.430 m/s * cos(45°)
vCE,y = 0.430 m/s * sin(45°)
The velocity of the river relative to the earth (vRE) only has an eastward component:
vRE,x = 0.760 m/s
Now, we can subtract the components to find the velocity of the canoe relative to the river:
vCR,x = vCE,x - vRE,x
vCR,y = vCE,y
To find the magnitude of vCR, we use the Pythagorean theorem:
|vCR| = sqrt(vCR,x^2 + vCR,y^2)
Substituting the given values:
vCR,x = 0.430 m/s * cos(45°) - 0.760 m/s
vCR,y = 0.430 m/s * sin(45°)
|vCR| = sqrt((0.430 m/s * cos(45°) - 0.760 m/s)^2 + (0.430 m/s * sin(45°))^2)
|vCR| ≈ 0.665 m/s
Therefore, the magnitude of the velocity vCR of the canoe relative to the river is approximately 0.665 m/s.
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Atmospheric pressure air at 100°C enters a 0.04m diameter tube.
and 2 m long with a speed of 9 m/s. A 1 kW electric heater
coiled on the outer surface of the tube provides a flow of heat
uniform to the tube. Determine: (a) the medical air flow rate, (b) the
air outlet temperature and (c) the tube wall temperature at
the exit
To solve this problem, we can use the principles of conservation of mass and energy.
(a) The mass flow rate of the air can be calculated using the formula:
mass flow rate = density * velocity * cross-sectional area
First, we need to determine the density of the air at 100°C. We can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Assuming atmospheric pressure at 100°C, we can calculate the density (ρ) using the equation:
ρ = P / (RT)
Substituting the values into the equation, we can calculate the density.
Once we have the density, we can calculate the mass flow rate using the given velocity and cross-sectional area.
(b) To determine the air outlet temperature, we can use the energy conservation equation:
mass flow rate * specific heat capacity * (T_out - T_in) = heat input
We know the mass flow rate from part (a), and the specific heat capacity of air can be looked up or assumed. The heat input is given as 1 kW.
Solving for T_out will give us the air outlet temperature.
(c) To determine the tube wall temperature at the exit, we need to consider the heat transfer from the heater to the air and the heat transfer from the air to the tube wall. This will depend on the thermal conductivity and the convective heat transfer coefficients.
Additional information about the thermal conductivity and convective heat transfer coefficients is needed to calculate the tube wall temperature accurately.
It's important to note that this problem requires more specific information about the properties of the tube, such as thermal conductivity and convective heat transfer coefficients, to provide an accurate solution.
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After 30 min of running, at the 9.0 km point in a 10 km race, you find yourself 150 m behind the leader and moving at the same speed as she does. What should your acceleration be if you're to catch up by the finish line? Assume that the leader maintains constant speed the entire race.
Express your answer with the appropriate units.
*I got 0.007 and it was wrong so please also give with correct sig figs
To catch up to the leader by the finish line, your acceleration should be approximately 0.356 m/s².
To catch up to the leader by the finish line, you need to cover the remaining 1 km while closing the 150 m gap. This means your total displacement needs to be 1.15 km (or 1150 m).
Given that you have 30 min (or 1800 s) remaining to cover this distance, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s is the displacement (1150 m)
u is the initial velocity (same as the leader's velocity, since you are moving at the same speed)
t is the time (1800 s)
a is the acceleration (what we need to find)
Rearranging the equation to solve for acceleration:
a = 2(s - ut) / t^2
Substituting the known values:
a = 2(1150 m - 0 m/s * 1800 s) / (1800 s)^2
Calculating the result:
a ≈ 0.356 m/s²
Therefore, your acceleration should be approximately 0.356 m/s² in order to catch up to the leader by the finish line.
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What is the wavelength of a wave with a velocity of 5.28 m/s and a frequency of 7.39 Hz ? 39.0 Hz. 2.11 Hz 1.40 Hz 0.714 Hz What is the name of the point E ? wavelength equilibrium position trough crest A string that is 0.250 m long has a bob with a mass of 12.3 amplitude is 0.100 m. What is the period of oscillation? 246 s 0.160 s 39.3 s 1.00 s The relative intensity of the threshold of hearing, 10
−12
W/m
2
, is 120 dB. −120 dB 0 dB −12 dB
Previous question
The wavelength of a wave with a velocity of 5.28 m/s and a frequency of 7.39 Hz is 0.715 m (rounded to three decimal places). The formula to find the wavelength is:λ = v/f,where λ is the wavelength, v is the velocity, and f is the frequency.Substituting the given values:λ = v/f = 5.28/7.39 = 0.715 m.
The name of the point E is the equilibrium position. This is the position where the object is at rest. In a simple harmonic motion, the equilibrium position is the position where the object oscillates about. The amplitude is the distance from the equilibrium position to the maximum displacement of the object.
The trough and crest are the points where the wave crosses the equilibrium position.The period of oscillation of a 0.250 m long string with a bob of mass 12.3 and amplitude 0.100 m is 0.160 s. The formula to find the period is:T = 2π√(L/g),where T is the period, L is the length of the string, and g is the acceleration due to gravity.Substituting the given values:T = 2π√(L/g) = 2π√(0.250/9.81) = 0.160 sThe relative intensity of the threshold of hearing, 10-12 W/m2, is 0 dB. The decibel (dB) is a logarithmic unit used to measure sound intensity. It is based on a ratio of the intensity of the sound to a reference level. In this case, the reference level is the threshold of hearing, which is 10-12 W/m2.
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