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Helga [31]
3 years ago
15

How much work is required to lift an object with a mass of 5.0 kilograms to a height of 3.5 meters?

Physics
2 answers:
KATRIN_1 [288]3 years ago
7 0
Work is Force multiplied by distance: work = Fd

work = mgh

work = (5.0 kg)(9.8 m/s)(3.5 m)

work = 171.5 joules or for two sigfigs, 170 joules
Veronika [31]3 years ago
6 0
I believe it would be 172 J (or 1.72 · 10² J)
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Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is
andrey2020 [161]
Considering that the book is moving with constant speed, the force applied by Anna must be the same that the friction force:

F = F_R = F\cdot \mu_k\cdot N

If we clear the previous equation:

\mu_k = \frac{F}{F_R} = \frac{19.5\ N}{51.7\ N} = \bf 0.38
5 0
2 years ago
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A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the
Ne4ueva [31]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8 
<span>the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance </span>

<span>the max height one is d=0.5*9.8*t^2 </span>
<span>d= max height subtract 1800-d</span>
3 0
2 years ago
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The acceleration due to gravity on the Moon's surface is
Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

g'=\frac{g}{6}

We know that weight of an object on Earth is,

w = m\times g

m = \frac{w}{g}

Similarly, weight on Moon will be

w' = m\times g'

w' = \frac{w}{g}\times\frac{g}{6}

w' = \frac{300}{6}

w' = 50

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0
3 years ago
What is used to block the UV light during screen development process to create a stencil in the emulsion
coldgirl [10]

Answer:

Invisible UV energy reacts with emulsion sensitizer and hardens the stencil so it won't dissolve with water and rinse down the drain

Explanation:

4 0
1 year ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
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