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3 years ago

5

A boulder with a mass of 50 kg, sits on a cliff that is 50 m high. The boulder is pushed off of the cliff. How much kinetic ener

gy does the boulder have just before it hits the ground?

1 answer:

rosijanka [135]3 years ago

8 0

50 +50 =100 Since it’s sitting on a 50m cliff that’s high with a mass of 50 kg it would be adding because once it goes down it’s adding speed

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Five different forces act on an object. Is it possible for the net force on the object to be zero?

bazaltina [42]

No because there must be an even # if their is an even amount one of the forces isn’t being cancelled

4 0

3 years ago

Sphere is fired downwards into a medium with an initial speed of 27 m/s. If it experiences a

anygoal [31]

Answer:

33 m/s

first i do 27+6=33 +s2=

8 0

3 years ago

a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp

stiks02 [169]

Answer:

1 / 2 m v^2 = L m g (1 - cos θ)

This is the KE due to the pendulum falling from a 25 deg displacement

v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2

v = 1.92 m/s this is the speed due to an initial displacement of 25 deg

Its speed at the bottom would then be

1.92 + 1.2 = 3.12 m/s since it gains 1.92 m/s from its initial displacement

3 0

1 year ago

A force of 32.5N stretches a spring of 0.500cm. What force will stretch a similar spring 1.60 cm?

Vlad [161]

Answer:

104 N

Explanation:

Calculate the spring stiffness:

F = kx

32.5 N = k (0.500 cm)

k = 65 N/cm

Find the force for the new length:

F = kx

F = (65 N/cm) (1.60 cm)

F = 104 N

3 0

3 years ago

For a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass 1000 kg. the truck then

babunello [35]

ANSWER

Both trucks will move together with speed v = 6.67 m/s

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

EXPLANATION

As we know that there is no external force on the system of two trucks

So here momentum of the two trucks before collision and after collision will remain same

So here we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

so here we will have

$v_{1i} = 10 m/s$

$v_{2i} = 0$

$m_1 = 2000 kg$

$m_2 = 1000 kg$

now we will have

$2000\times 10 + 1000\times 0 = (2000 + 1000) v$

$v = \frac{20000}{3000} = 6.67 m/s$

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

3 0

3 years ago

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