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gogolik [260]
11 months ago
13

A 2.0 kg book is lying on a 0.78 m -high table. You pick it up and place it on a bookshelf 2.1 m above the floor.Part ADuring th

is process, how much work does gravity do on the book?Part BDuring this process, how much work does your hand do on the book?
Physics
1 answer:
zloy xaker [14]11 months ago
7 0

Given data:

* The mass of the book is 2 kg.

* The initial height of the book is 0.78 m.

* The final height of the book is 2.1 m.

Solution:

(A). The work done by the gravity on the book is,

W=mg(h_f-h_i)

where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,

The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.

Substituting the known values,

\begin{gathered} W=2\times(-9.8)\times(2.1-0.78) \\ W=-19.6\times1.32 \\ W=-25.9\text{ J} \end{gathered}

Thus, the work done by the gravitational force is -25.9 J.

(B). The work done by the hand on the moving the book is,

\begin{gathered} W_1=-W \\ =25.9\text{ J} \end{gathered}

Thus, the work done by the hand on the book is 25.9 J.

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ch4aika [34]

Answer:

At 11.5 m

Explanation:

The power per unit area corresponds to the intensity, which is given by

I=\frac{P}{4\pi r^2}

where

P is the power

4\pi r^2 is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)

Here we want the intensity of the two light bulbs to be the same, so

I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}

where we have

P1 = 100 W is the power of the first light bulb

P2 = 75 W is the power of the second light bulb

r2 = 10 m is the distance from the second light bulb

Solving for r1, we find

r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m

4 0
3 years ago
Planet Tatoone is about 1.7 AU from its Sun. Approximately how long will it take for light to travel from the Sun to Tatoone in
Radda [10]

Answer:

The value is   t =  14.129 \  minutes    

Explanation:

From the question we are told that

  The distance of planet Tatoone is  d =  1.7 \ AU  =  1.7 *1.496* 10^{11}=2.543*10^{11} \ m

   The  speed of light is  c =  3.0*10^{8} \  m/  s

Generally the time taken is mathematically represented as

     t =  \frac{d}{c}

=> t =  \frac{2.543*10^{11}}{3.0*10^{8} }

=>    t =  847.7 \  s

Now converting to minutes

       t =  \frac{847.7}{60}

   =>     t =  14.129 \  minutes    

8 0
3 years ago
A car drives to the east in a time of 4 hours. Then, immediately (not realistic, but just assume this is the case for this probl
diamong [38]

Answer:

Average speed in east direction speed=\frac{distance}{time}=\frac{28}{4}=7km/hr            

Explanation:

We have given average speed of the entire trip = 5 km/hr

First the car 4 hours then travels 12 km in 4 hours

So total time of the trip = 4+4 = 8 hours

So total distance traveled in the trip d=speed\times time = 5\times 8=40km

As the car travel 12 km in 4 hours to the west

So distance traveled in 4 hour in east = 40 -12 = 28 km

Time in east direction = 4 hour

So average speed in east direction speed=\frac{distance}{time}=\frac{28}{4}=7km/hr

4 0
3 years ago
The repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?
Aloiza [94]

Answer:

The answer to your question is letter A.     r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

                 F = K\frac{q1q2}{r^{2}}

Solve for r

                r = \sqrt{\frac{kq1q2}{F}}

Substitution

                r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}

Simplification

                r = \sqrt{\frac{2.3 x 10^{-28}}{2}}

                r = \sqrt{1.15 x 10^{-24}}

Result

                r = 1.07 x 10⁻¹⁴ m

6 0
3 years ago
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pychu [463]
Hello!!

Here we have a simple matter of conservation of energy. ME=PE+KE.

At point A we have PE=mgh and KE=1/2mv^2. At point A all we have is PE since the coaster isn’t rolling yet. But by conservation of energy, we know that it will have enough energy to roll down and get to and equal height on another hill. Providing we are neglecting friction and drag and resistance forces which we are in this case. So we can conclude that the KE will be greater at Point B since ME=PE+KE and for ME to remain the same and we know the PE is less on lower hill, so we can conclude that KE on lower hill is greater to keep ME the same and have conservation of energy.

Hope this helps you understand the concept!! Any questions please just ask!! Thank you so much!!
7 0
3 years ago
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