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Kaylis [27]
4 years ago
9

Power can be defined as the distance over which work was done. how much work can be done in a given time. all the work in a give

n area. the energy required to do work.
Physics
1 answer:
Artist 52 [7]4 years ago
4 0

Power is defined as rate of work done or rate of energy

so here we can say that

Power = \frac{Work}{time}

so here correct answer must be

how much work can be done in a given time.

So we need to find the work done in a given period of time and then we will have to find the ratio of total work done and total time taken.

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Which best compares kinetic energy and temperature?
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Kinetic energy is energy of motion while temperature is a measure of that energy in substances.
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3 years ago
Read 2 more answers
A stone is thrown vertically upwards with a speed of 30.0 m/s.
matrenka [14]

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s


b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m



6 0
3 years ago
Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
Drupady [299]

Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

F_K=u_k*m*g

u_K*m*g=m*a_c

u_K*g=a

u_K=\frac{a_c}{g}

a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}

u_K=0.862

4 0
3 years ago
Manipulate the equation "v=d/t" to find the answers to these problems using
OLEGan [10]
Yo no me voy a ir a la cama a
4 0
3 years ago
A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What
Jobisdone [24]

Answer:

v_{avg} = 6.41 m/s

Explanation:

Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

here we know that initially the sky diver drops without opening parachute by total displacement 625 m

then she open her parachute and drop another 362 m

so first it took time t = 15 s to drop without open parachute

then it took t = 139 s to drop next displacement

so here total displacement is given as

d = 625 m + 362 m

total time is given as

t = 15 s + 139 s

so average velocity is given as

v_{avg} = \frac{625 + 362}{15 + 139}

v_{avg} = 6.41 m/s

4 0
3 years ago
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