Ask question

Ask question

All categories

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

You might be interested in

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

The gravity is equal to 0.8, then the acceleration is

$a = 480*\frac{g}{9.8}$

$a = 53.3g$

3 0

3 years ago

Neritic Sediments are deposited on the ocean floor in a sorted manor. In which order are the sediments ordered moving from the s

BaLLatris [955]

B <span> of Earth’s surface is covered by water. Very little or no light penetrates beyond a few hundred feet in water</span>

6 0

3 years ago

What effect does the sun have on the earth. written response

maw [93]

It had a gravitational force that acts so that it pulls Earth in an orbital, which explains temperature variations throughout the years.

8 0

3 years ago

What is the netforce?

Kipish [7]

Answer:

6 Newtons to the left.

Explanation:

We can convert this into a generic algebra equation by giving directions positive and negative values.

The 6 will be positive, and the 10 and 2 will be negative.

Add 10 and 2 to have 12.

6-12 = -6.

Therefore you have 6 newtons to the left (negative).

3 0

3 years ago

Read 2 more answers

In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2 (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

3 years ago