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IceJOKER [234]
3 years ago
8

An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by

the force on the box?
a. 6.0 × 101 joules
b. -6.0 × 101 joules
c. 5.9 × 102 joules
d. -5.9 × 102 joules
Physics
1 answer:
devlian [24]3 years ago
3 0

Answer:

W ≈ 5.9 × 10^{2} J

Hence, option (c) is correct.

Given:

mass of box = 6 kg

Displacement of box = 10 m

To find:

Work done by the force = ?

Formula used:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

Solution:

Work done is equal to change is potential energy,

W = m g h

Where, W = Work done

m = mass of the box

h = displacement by the force

W = 6 × 9.8 ×10

W = 588 Joule

Thus, W ≈ 5.9 × 10^{2} J

Hence, option (c) is correct.

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<h2>First stage of sleep Deprivation Subject </h2>

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7 0
2 years ago
Read 2 more answers
One factor that could account for a drop in the water table is:
Shtirlitz [24]

Answer:

D. Less rain and snow.  

Explanation:

A factor that can be a account for a drop in the water table is, less rain and snow. to topography, water tables is influenced by lot of factors, including the geology, weather, ground cover.

4 0
2 years ago
For a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass 1000 kg. the truck then
babunello [35]

ANSWER

Both trucks will move together with speed v = 6.67 m/s

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

EXPLANATION

As we know that there is no external force on the system of two trucks

So here momentum of the two trucks before collision and after collision will remain same

So here we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

so here we will have

v_{1i} = 10 m/s

v_{2i} = 0

m_1 = 2000 kg

m_2 = 1000 kg

now we will have

2000\times 10 + 1000\times 0 = (2000 + 1000) v

v = \frac{20000}{3000} = 6.67 m/s

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

3 0
2 years ago
Read 2 more answers
magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,
Mama L [17]

Answer:

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

Explanation:

Given:

height above which the rock is thrown up, \Delta h=50\ m

initial velocity of projection, u=20\ m.s^{-1}

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

v=u+g'.t'

where:

v= final velocity at the top height = 0 m.s^{-1}

0=20-g'.t' (-ve sign to indicate that acceleration acts opposite to the velocity)

t'=\frac{20}{g'}\ s

The time taken by the rock to reach the top height on the earth:

v=u+g.t

0=20-g.t

t=\frac{20}{g} \ s

Height reached by the rock above the point of throwing on the exoplanet:

v^2=u^2+2g'.h'

where:

v= final velocity at the top height = 0 m.s^{-1}

0^2=20^2-2\times g'.h'

h'=\frac{200}{g'}\ m

Height reached by the rock above the point of throwing on the earth:

v^2=u^2+2g.h

0^2=20^2-2g.h

h=\frac{200}{g}\ m

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2 (during falling it falls below the cliff)

here:

u= initial velocity= 0 m.s^{-1}

\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2

t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}

t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }

Similarly on earth:

t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g}  }

Now the required time difference:

\Delta t=(t'+t_f')-(t+t_f)

\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'}  }  )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g}  }  )

3 0
3 years ago
A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20, what is the exact distance that the seismic station is away from
maxonik [38]

The exact distance from the seismic station to the epicenter is 6000 km.

<h3>Epicenter of earthquake</h3>

The earthquake's epicenter is the point above the fault location on the earth's surface.

Given that:

A p-wave arrives at 3:00:00 and the s-wave arrives at 3:07:20.

Difference in arrival time = 3:07:20 - 3:00:00 = 7 minutes 20 seconds

From the earthquake time travel graph, a time difference of 7 min 20 sec is at x = 6

The exact distance from the seismic station to the epicenter is 6000 km.

Find out more on Epicenter of earthquake at: brainly.com/question/1969968

4 0
2 years ago
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