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VARVARA [1.3K]
3 years ago
15

How fast an object is moving is its

Physics
1 answer:
faltersainse [42]3 years ago
4 0

Answer:

D. Speed !!!!!

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A box of mass m is pushed at an angle 0 by a force Fp along a frictionless surface. The box travels to right with acceleration a
Genrish500 [490]

Answer:A block rests on a horizontal, frictionless surface. A string is attached to the block, and is pulled with a force of 45.0 N at an…

Explanation:

5 0
3 years ago
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Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in
Elanso [62]

Answer:

kinetic energy

Explanation:

a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.

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What velocity does a 2kg mass have when its kinetic energy is 16 J
alexandr402 [8]
We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v. 

K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s

Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!
7 0
3 years ago
A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision
Nimfa-mama [501]

Answer:

Δ KE =  249158.6 kJ  

Explanation:

given data

Truck mass  M =  1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum  that is express as

Mu = (M+m) V     .......................1

put here  value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy  will be here

Δ KE =   \frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2         .................2

put here value and we get

 Δ KE = \frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2  

solve it we get

Δ KE =  249158.6 kJ  

6 0
3 years ago
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