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Naily [24]
3 years ago
10

A man applies a force of 100 Newtons to a rock for 60 seconds, but the rock does not not move. What is the amount of work done b

y the man on the rock
Physics
1 answer:
frozen [14]3 years ago
7 0
Well if the rock doesn't move, then there is no amount of work done. There is no work done on an object if a force is applied to the object but it DOES NOT change its position, in this case is the rock.
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A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me
enyata [817]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

6 0
2 years ago
An object at rest starts accelerating.
Shalnov [3]

Answer:

<u>We are given: </u>

initial velocity (u) = 0 m/s

final velocity (v) = 10 m/s

displacement (s) = 20 m

acceleration (a) = a m/s/s

<u>Solving for 'a'</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(10)² - (0)² = 2(a)(20)

100 = 40a

a = 100 / 40

a = 2.5 m/s²

6 0
3 years ago
Indiana jones (83.5 kg) is running 3.75 m/s when he jumps in a stationary 312 kg mine cart. what is their joint velocity afterwa
Lubov Fominskaja [6]

Answer:

.7917 m/s

Explanation:

This is a conservation of momentum question. You have an object initially at rest (cart) so that object is initially at 0 momentum. Indiana Jones is 83.5 kg and running 3.75 m/s so he starts with a momentum of 313.125 kg * m/s because momentum is equal to mass * velocity. Once the person jumps in the cart, the cart and the person can be considered one object and by conservation of momentum, the momentum of the Indiana-cart system is equal to 313.125 kg * m/s. By that, we can set that momentum equal to the combined mass * joint velocity. So 313.125 = (83.5kg + 312kg) * joint velocity. Then just solve for the velocity. The answer should be smaller than the intial velocity of the person of 3.75 m/s because the mine cart is HUGE at 312kg.

3 0
3 years ago
The sphere that refers to Earths water is called the
soldier1979 [14.2K]
Earth Spheres. Earth's Spheres. Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called “spheres.” Specifically, they are the lithosphere (land), hydrosphere (water), biosphere (living things), and atmosphere (air).
6 0
3 years ago
Read 2 more answers
An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
3 years ago
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