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3 years ago

Describe the mechanical energy and the thermal energy of a bouncing basketball

Physics

1 answer:

miskamm [114]3 years ago

8 0

Answer:

when the ball is bounced gravitationally energy is released (more info below)

Explanation:

When a force is applied down to the ball, the gravitational potential energy is converted into kinetic energy before it hits the earth. Cinetic energy is converted into vibration, thermal, and static electrical energy, before the momentum of the ball pushes the ball back up into the hands of the people.

hope this helped!

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Did the strongman pull up or push down to lift the pig

fredd [130]

Answer:

pull up???????????????

7 0

3 years ago

Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.4

Free_Kalibri [48]

Answer:

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

$\alpha$ =6.125 rad/ $s^{2}$

Tension on side A = 4.5 × g= 44.1 m/ $s^{2}$

Tension on side B= 2.0 × g= 19.6 m/ $s^{2}$

Explanation:

As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.

Observe the figure attached, let the tension with Block A be $T_{2}$ and the tension attached with Block B be $T_{1}$ .

Tensions will be only be due to the weight of the blocks as no other force is present.

$T_{2}$ = 4.5 × g= 44.1 m/ $s^{2}$

$T_{1}$ = 2.0 × g= 19.6 m/ $s^{2}$

Now, lets make a torque equation about the center of the wheel and find the alpha

$T_{2}$ ×R- $T_{1}$ ×R= MI( Moment of Inertia of Wheel)× Alpha

where, R= Radius of the wheel=0.100m and

Alpha( $\alpha$ )= Angular acceleration of the wheel

MI of the wheel= 0.400 kg/ $m^{2}$

$(44.1-19.6)R=0.400\alpha$

$\alpha = \frac{24.5 * 0.100}{0.400}$

$\alpha$ =6.125 rad/ $s^{2}$

Acceleration = R × $\alpha$

= 0.1 * 6.125

=0.6125 $m/s^{2}$

Accelerations of both the sides is 0.6125 $m/s^{2}$ , A moves downwards whereas B moves upwards.

7 0

3 years ago

How can u prove that there is an electric field

zimovet [89]

Explanation:

By introducing a particle of one positive coulomb between the electric plates, the particle experience a force against the electrostatic forces, proving that it is in an electric field

The force experienced by this particle is known as Electric Field Intensity  denoted by E and it is given by

$k \times \frac{q {}^{2} }{r}$

where, k is constant =

$9 \times {10}^{9}$

q is the charge magnitude on the particle

r is the separation of the particles

6 0

3 years ago

A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h ove

Phantasy [73]

Answer:

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

5 0

3 years ago

A heavy-duty stapling gun uses a 0.131-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff sprin

azamat

Answer:

v = 15.3 m/s

Explanation:

Given:

- The mass of the metal rod m = 0.131 kg

- The spring constant of spring k = 35666 N / m

- The initial compression of the spring x_i = 3.2 ✕ 10^-2m

- The final compression of the spring x_f = 1.3 ✕ 10^-2m

Find:

Find the speed of the ram at the instant of contact.

Solution:

- The ram is initially at rest hence its initial velocity is zero. From conservation of energy principle we have and exchange of gravitational potential and elastic potential energy stored in the spring at initial and final state points:

0.5*m*v^2 + 0.5*k*x_2^2 + m*g*h_f= 0.5*k*x_1^2 + m*g*h_o

m*v^2 + k*x_2^2 + 2m*g*h_f = k*x_1^2 + 2*m*g*h_o

m*v^2 = k*x_1^2 - k*x_2^2 + 2*m*g*(h_o - h_f)

v^2 = k/m*[x_1^2 - x_2^2] + 2*g*(h_o - h_f)

v = sqrt { k/m*[x_1^2 - x_2^2] + 2*g*(h_o - h_f) }

- Where, h_o: is the initial potential energy at compression state x_1.

h_f : is the initial potential energy at compression state x_2

- Plug in the values:

v = sqrt { 35666/.131*[0.032^2 - 0.013^2] + 2*g*(0.032 - 0.013) }

v = sqrt { 232.7819 + 0.37278 } = sqrt { 233.1568 }

v = 15.3 m/s

3 0

3 years ago