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lora16 [44]
3 years ago
9

Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?

Physics
2 answers:
balandron [24]3 years ago
8 0

Answer:

The normal force that the wall exerts on the ball.

Explanation:

Erica throws a tennis ball against a wall, and it bounces back. Here Newton's third law of motion follows. When the ball hits the wall, the wall will also exert an equal and opposite force on the ball as per third law. The opposite force that wall exert on ball is normal force which is responsible for sending the ball back to Erica.

Hence, the force is normal force exerted by the wall on the ball.

matrenka [14]3 years ago
6 0
Kinetic energy is responsible for this

You might be interested in
Type of force that holds the nucleus of an atom together
Kamila [148]

Answer:

Nuclear Forces

Explanation:

The type  of force that holds the nucleus of an atom together is called Nuclear Forces.

7 0
3 years ago
Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun
Dominik [7]

Answer:

a. K_{Axis}=2.574x10^{29}J

b. K_{Orbit}=2.6577x10^{33}J

Explanation:

K_{Axis}=\frac{1}{2}I*w^2

I_{Sphere}=\frac{2}{5}*m*r^2

w=\frac{2\pi }{T} , T=24hrs*\frac{3600s}{1hr} =86400s

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2

K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2

K_{Axis}=2.574x10^{29}J

b.

T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s

K_{Orbit}=\frac{1}{2}*I*w

I=m*r^2

K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2

K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2

K_{Orbit}=2.6577x10^{33}J

4 0
3 years ago
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h
Alex73 [517]

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

5 0
3 years ago
Two students were pushing a heavy sofa. One student pushed with a force of 200.0 N to the right, while the other pushed with a f
makvit [3.9K]

Answer:

Explanation:

F = ma

a = F/m

a = (200.0 + 150.0 - 100.0) / 91.0

a = 250.0/91.0

a = 2.7472527...

a = 2.75 m/s²

8 0
2 years ago
A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .if after the collision it moves with a velocity of 2
ch4aika [34]

Answer:

-1.2 kg - m/s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .}\\\textsf{ After the collision it moves with a velocity of 2.0m/s in the opposite direction.}\end{cases}

And we need to find out the change in momentum of the body . Here ,

  • velocity before collision (u) = 10m/s
  • velocity after collision (v) = 2m/s .

We know that momentum is defined as amount of motion contained in a body . Mathematically ,

\sf\longrightarrow momentum (p)= mass(m) * velocity(v)

Therefore change in momentum will be,

\sf\longrightarrow \triangle p = mv - mu

Since the direction of velocity changes after the collision , the velocity will be -2m/s .

\sf\longrightarrow \Delta p = 100g( -2m/s -10m/s) \\

\sf\longrightarrow \Delta p =\dfrac{100}{1000}kg ( -12m/s)  \\

\sf\longrightarrow \Delta p   = 0.1 kg * -12m/s \\

\sf\longrightarrow \boxed{\bf \Delta p = -1.2 \ kg-m/s} \\

7 0
2 years ago
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