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yanalaym [24]
2 years ago
5

You pick up a 3.4 kg can of paint from the ground and lift it to a height of 1.8 m. how much work do you do on the can of paint?

you hold the can stationary for half a minute, waiting for a friend on a ladder to take it. how much work do you do during this time? you friend decides against the paint, so you lower it back to the ground. how much work do you do on the can as you lower it?
Physics
1 answer:
zepelin [54]2 years ago
5 0

1) +60.0 J

The work done when lifting an object is equal to the change in gravitational potential energy of the object:

W=\Delta U= m g \Delta h

where m is the mass of the object, g is the gravitational acceleration, \Delta h is the variation of height of the object.

In this problem, m = 3.4 kg, g = 9.8 m/s^2 and \Delta h = +1.8 m, so the work done is:

W=(3.4 kg)(9.8 m/s^2)(1.8 m)=+60.0 J

And the work done is positive, since the object has gained potential energy.


2) 0 J

In this case, there is no variation of the heigth of the object:

\Delta h =0

Therefore, the variation of potential energy is zero:

\Delta U = 0

And so, the work done is zero as well:

W=0


3) -60.0 J

In this case, the variation of height of the object is negative, since the object has been lowered to the ground:

\Delta h = -1.8 m

So, the object has lost potential energy:

W=(3.4 kg)(9.8 m/s^2)(-1.8 m)=-60.0 J

And so the work done is negative:

W=-60.0 J

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Bas_tet [7]
C is the answer hope that helps you
3 0
3 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
2 years ago
The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud
motikmotik
<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = \frac{Gm_1m_2}{r^2}

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = \sqrt{\frac{Gm_1m_2}{F} }

If we substitute the values in the above equation

r = \sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }

= 2.46 m

3 0
3 years ago
Read 2 more answers
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experimen
Nikolay [14]

Answer:

a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center

Explanation:

Let  n₁ and n₂ be no of lines per unit length  of grating A and B respectively.

λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,

Distance of first maxima for green light

= λ₁ D/ d₁

Distance of first maxima for red light

= λ₂ D/ d₂

Given that

λ₁ D/ d₁ = λ₂ D/ d₂

λ₁ / d₁ = λ₂ / d₂

λ₁ / λ₂  = d₁ / d₂

But

λ₁  <  λ₂

d₁ < d₂

Therefore no of lines per unit length of grating A will be more because

no of lines per unit length  ∝ 1 / d

If grating B is illuminated with green light first maxima will be at distance

λ₁ D/ d₂

As λ₁ < λ₂

λ₁ D/ d₂ < λ₂ D/ d₂

λ₁ D/ d₂ < 1 m

In this case position of first maxima will be less than 1 meter.

Option a is correct .

5 0
3 years ago
Can someone please help me
guajiro [1.7K]
ANSWER:
C. Small, minimize

Hope it helps u!
5 0
3 years ago
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