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Kitty [74]
2 years ago
12

o illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using t

he work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone.
Physics
1 answer:
Phantasy [73]2 years ago
5 0

Answer: force of gravity on the body due to height difference above the earth's surface

Explanation: as you increase the height of a body above ground, you do work against gravity in moving it from a point on the earth's surface to that point. So a body falling has a stored up gravito-potential energy which acts on it downward due to its mass, accelerating it downwards

Answer b): kinetic energy of the body

Explanation: the downward force produces an acceleration of magnitude 9.81m/s2 downwards which means an increasing velocity. This increasing velocity means the kinetic energy of the body is increasing (kinetic energy is proportional to velocity of the body squared)

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Hmm.. Well i would say A or C because water and acid would just soak threw it

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Instructions:Drag the tiles to the correct boxes to complete the pairs. Match each term with its definition. Tiles conductor rad
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These are the answers:

</span>Conductor - <span>substance that allows heat through 
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1 year ago
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A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
2 years ago
A spring-powered dart gun is unstretched and has a spring constant 16.0 N/m. The spring is compressed by 8.0 cm and a 5.0 gram p
stepladder [879]

Answer:

Explanation:

Given that,

Spring constant = 16N/m

Extension of spring

x = 8cm = 0.08m

Mass

m = 5g =5/1000 = 0.005 kg

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

So, Using conservation of energy

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4 0
1 year ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

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h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

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13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
1 year ago
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