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Sphinxa [80]
3 years ago
9

A box with a mass of 7 kilograms is pushed up a ramp to a height of 5 meters. What is the work done against the force of gravity

? Assume the ramp has negligible friction.
A. 3.4 × 102 joules

B. 6.9 × 102 joules

C. 8.6 × 102 joules

D. 1.7 × 102 joules
Physics
2 answers:
zloy xaker [14]3 years ago
7 0
A. 3.4 x 10 2 its right i guessed it
Musya8 [376]3 years ago
4 0

The correct answer to the question is 3.4\times 10^2\ J .

CALCULATION:

As per the question, the mass of the box is given as m = 7 kg.

The box was pushed to a height h = 5 m.

We are asked to calculate the work done by the box.

We know that energy is the capacity to do work.The work done by a body when it is taken to certain height above the surface of earth against gravity is equal to the potential energy gained by that body .

Hence, the work done W = P.E

The potential energy gained by the box when it was taken to certain height is the gravitational potential energy of the box.

The gravitational potential energy gained by the box is calculated as -

                      Potential energy P.E = mgh

                                                         = 7 × 9.8 × 5 J.

                                                         = 343 J

                                                         = 3.43\times 10^2\ J

                                                         = 3.4\times 10^2\ J

Here, g is known as the acceleration due to gravity and g = 9.8\ m/s^2

We know that work done by the box against gravity W = P.E.

Hence, the work done by the box against gravity is 3.4\times 10^2\ J

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Explanation: Hope that helps broski

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Answer:

T = 99.51 hour

Explanation:

Mass of Uranus, M=8.68\times 10^{25}\ kg

The moon Umbriel orbits Uranus at a distance of 2.66\times 10^8\ m

We need to find Umbriel's orbital period. Let it is T. Using Kepler's third law of motion to find it.

T^2\propto r^3\\\\T^2=\dfrac{4\pi^2r^3}{GM}\\\\T^2=\dfrac{4\pi^2\times (2.66\times 10^8)^3}{6.67\times 10^{-11}\times 8.68\times 10^{25}}\\\\T=358244.51\ s

As 1 hour = 3600 s

358244.51 s = 99.51 hour

Hence, Umbriel's orbital period is 99.51 hour.

7 0
3 years ago
The plates on a capacitor have a radius of 1.5 mm and are separated by a distance of 0.43 mm. The space between the plates is fi
zhenek [66]
So, C = kE°A/d

putting the values,

C
= 3.8 × 8.85×10^(-12) × 3.14×1.5×1.5 × 10^(-6)/0.43 × 10^(-3)

so, 1.02 × 10^(-13)

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A solid of density 8000 kgm.. weighs 0.8 kgf in air. When it is completely
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Explanation:

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m1= mass 1 = 1.1 kg

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m2= 2.4 kg

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We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2

Replacing:

Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9Vf1=-1-2.6=-3.6\text{ m/s}

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6 0
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