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mihalych1998 [28]
3 years ago
6

A car possesses 20 000 units of momentum. What would be the car's new momentum if its mass was doubled

Physics
1 answer:
Klio2033 [76]3 years ago
7 0

Answer:

40 000 units of momentum

Explanation:

Assuming that the mass is doubled and the velocity stays constant the new momentum will be doubled the original momentum.

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A ball is thrown up into the air with 100 j of kinetic energy, which is transformed to gravitational potential energy at the top
jenyasd209 [6]

The kinetic energy when it returns to its original height is 100 J

Solution:

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

<u>u² = v² -2·g·s</u>

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

As the height of the ball increases the kinetic energy of the ball is converted into gravitational potential energy. This means that the kinetic energy of the bullet is reduced. When the ball reaches its maximum height, it momentarily comes to rest and the ball's kinetic energy is zero. When the ball hits the ground, its potential energy is converted to kinetic energy.

Learn more about Kinetic energy here:-brainly.com/question/25959744

#SPJ4

6 0
1 year ago
A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr
Likurg_2 [28]

Answer:

Part a)

T = 0.81 s

Part b)

v_x = 3.33 m/s

Part c)

v_y = 3.91 m/s

Part d)

\theta = 49.55 degree

Part e)

T = 1.11 s

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

\Delta y = 1.80 - 1.02

\Delta y = 0.78 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(0.78)

v_y = 3.91 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 3.91 - 9.81 t_1

t_1 = 0.39 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(1.80 - 0.95)

v_y = 4.08 m/s

time taken by it to reach this height

4.08 = v_i + at

4.08 = 0 + 9.81 t_2

t_2 = 0.42 s

T = t_1 + t_2

T = 0.81 s

Part b)

Horizontal velocity

v_x = \frac{x}{t}

v_x = \frac{2.70}{0.81}

v_x = 3.33 m/s

Part c)

vertical velocity is the intial y direction velocity

v_y = 3.91 m/s

Part d)

Take off angle is given as

tan\theta = \frac{3.91}{3.33}

\theta = 49.55 degree

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

\Delta y = 2.50 - 1.20

\Delta y = 1.30 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(1.30)

v_y = 5.05 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 5.05 - 9.81 t_1

t_1 = 0.51 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(2.50 - 0.72)

v_y = 5.9 m/s

time taken by it to reach this height

5.9 = v_i + at

5.9 = 0 + 9.81 t_2

t_2 = 0.60 s

T = t_1 + t_2

T = 1.11 s

5 0
2 years ago
Which way will the people on the right move? Why? Plz answer
Scrat [10]
Probably to the right because the force on the left is greater and will therefore overpower the 2N force and push to the right
3 0
3 years ago
5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How
ad-work [718]

Answer:

60 m/s

Explanation:

a =  \frac{vf - vi}{t}

5.0 =  \frac{vf - 0}{12}

5.0 \times 12 = vf

60 = vf

6 0
3 years ago
b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force
Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley  = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

Torque is

= force \times  radius

= f \times r

And,

Torque is also

= moment\ of\ inertia \times angular\ acceleration

= I \times \alpha

So,

We can say that

f \times r = I \times \alpha

f \times 0.05 = 0.2 \times 2

0.05f = 0.4

f = 8 N

We simply applied the above formulas

8 0
3 years ago
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