An object at free fall near the Earth’s surface experiences constant velocity.

A 2 kg soccer ball is traveling 28.62m/s when it hits the wall and bounces off of the wall with a velocity of 20 m/s. If the wal

Lesechka [4]

Answer:

0.149 s or 0.15 s

Explanation:

let initially ball is moving towards left hence initial velocity = - 28.62 m/s

final velocity as ball moves right = +20 m/s

force = rate of change in momentum

force = mass × change in velocity / time

or time = mass × change in velocity / force

time = 2× ( 20 -( -28.62)) / 652.36

time = 2× ( 20 +28.62)) / 652.36

time = 2× 48 .62/652.36

time = 0.149 s or 0.15 s

5 0

4 years ago

What is the energy dissipated as a function of time in a circular loop of N turns of wire having a radius of r0 and a resistance

Snezhnost [94]

Explanation:

Let N be the number of turns in a circular loop having a radius of r₀ and a resistance R. The magnetic field is given by :

$B(t)=B_oe^{-\dfrac{t}{\tau}}$

The induced emf in the circular coil is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi=BA$

$\epsilon=N\dfrac{-d(BA)}{dt}$

$\epsilon=NAB_o\dfrac{-d(e^{-\dfrac{t}{\tau}})}{dt}$

$\epsilon=\dfrac{NAB_o}{\tau}e^{-t/\tau}$

Power dissipated is given by :

$P=\dfrac{\epsilon^2}{R}$

$P=\dfrac{(\dfrac{NAB_o}{\tau}e^{-t/\tau})^2}{R}$

$P=(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}$

Energy dissipated in the circuit is given by :

$E=\int\limits^T_0 {P.dt}$

$E=\int\limits^T_0 {(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}.dt}$

$E=\dfrac{1}{R}(\dfrac{NAB_o}{\tau})^2\int\limits^T_0 {e^{-2t/\tau}.dt}$

$E=\dfrac{(N\pi r_o^2B_o^2)^2}{2R}{\tau(e^{2t/\tau}-1)}{}$

Hence, this is the required solution.

7 0

3 years ago

PLEASE HELP QUICKLY!!

tensa zangetsu [6.8K]

Answer:

A

Explanation:

took the test today at school, got a 95.

6 0

3 years ago

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A train moves with a constant velocity for 15s for 155m. How fast is the train moving?

bonufazy [111]

Answer: 10.3m/s

Explanation:

In theory and for a constant velocity the physics expression states that:

Eq(1): distance = velocity times time <=> d = v*t for v=constant.

If we solve Eq (1) for the velocity (v) we obtain:

Eq(2): velocity = distance divided by time <=> v = d/t

Substituting the known values for t=15s and d=155m we get:

v = 155 / 15 <=> v = 10.3

5 0

3 years ago

Pleaswe help me ewwfea

Talja [164]

Answer:

The first one is Earths Tilt on its axis

6 0

3 years ago

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