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Dmitry [639]
1 year ago
10

kg book is lying ona 0.80 m -high table. You pick it up and placeit on a bookshelf 2.2 mabove the floor.a.During this process, h

ow much work does gravity do on thebook. Express your answer to two significant figuresand include the appropriate units.b. During this process, how much work doesyour hand do on the book?Express your answer to two significant figures and include theappropriate units.
Physics
1 answer:
Sergeu [11.5K]1 year ago
8 0

The work done by gravity is  43 J while the work done by your habd is 27 J.

<h3>What is the work done?</h3>

We know that all over the earth is a large gravitational field. If an object is placed at a given height, the kind of energy that the object would be said to possess at that height is the gravitational potential energy. This is the kind of energy that a body would have because it is found at a given height on the earth surface.

Now, we know that gravity does some work on the book as it found to be at a position of about 2.2 on top of the shelf. The work that is done by the force of gravity on the book is obtained as;

W = mgh

W = work done

m = mass

g = acceleration due to gravity

h = height

W = 2 * 9.8 * 2.2

= 43 J

The work that is done by your hand =

W = 2 * 9.8 (2.2 - 0.8)

= 27 J

Learn more about work done:brainly.com/question/10334435

#SPJ1

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8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
Dana is on a train traveling at a speed of 20 km/h. Dana walks from the front of the train to the back of the train at a speed o
Maslowich

Answer:

16km/h

Explanation:

Vt=20km/h ---train speed

Vd=4km/h

Donas speed relative to ground is:

Vrd=Vt-Vd

Donas is moving in opposite direction of train .

Vrd=20km/h-4km/h

Vrd=16km/h

7 0
3 years ago
A long, straight wire lies in the plane of a circular coil with a radius of 0.018 m. the wire carries a current of 5.6 a and is
iris [78.8K]
(a) The net flux through the coil is zero.
In fact, the magnetic field generated by the wire forms concentric circles around the wire. The wire is placed along the diameter of the coil, so we can imagine as it divides the  coil into two emisphere. Therefore, the magnetic field of the wire is perpendicular to the plane of the coil, but the direction of the field is opposite in the two emispheres. Since the two emispheres have same area, then the magnetic fluxes in the two emispheres are equal but opposite in sign, and so they cancel out when summing them together to find the net flux.

(b) If the wire passes through the center of the coil but it is perpendicular to the plane of the wire, the net flux through the coil is still zero.
In fact, the magnetic field generated by the wire forms concentric lines around the wire, so it is parallel to the plane of the coil. But the flux is equal to
\Phi = BA \cos \theta
where \theta is the angle between the direction of the magnetic field and the perpendicular to the plane of the coil, so in this case \theta=90^{\circ} and so the cosine is zero, therefore the net flux is zero.
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3 years ago
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7 0
3 years ago
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A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease
skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm

So, the new pressure is 2 atm.

7 0
2 years ago
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