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Nitella [24]
2 years ago
8

You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?​

Physics
2 answers:
77julia77 [94]2 years ago
5 0

Answer:

20 J of work is done by you, against gravity

Explanation:

Work done by gravity = Potential energy stored in the book after it is put ona shelf.

Potential energy stored:

U = mgh

U = wh

w= weight in Newton

m= mass of the book

g= acceleration due to gravity

h= height through which, it is raised.

Work is being done against gravity.

U = 10 \times 2

U = 20 J

Murljashka [212]2 years ago
3 0

Answer:

20 J

Explanation:

Given:

Weight of the book is, W=10\ N

Height or displacement of the book is, d=2\ m

The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.

So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.

Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.

Therefore, work done is given as:

Work=W\times d\\Work=10\times 2=20\ J

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.

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The temperature of the liquid in a container decreases as the liquid evaporates. Use kinetic theory to explain why.
melisa1 [442]

Answer:

Temperature decreases because the number of collision of the molecules decreases as they escape or evaporate. Molecules are in constant motion. Increase in temperature leads to increase in average kinetic energy of the molecules.

6 0
2 years ago
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2
LiRa [457]

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

6 0
2 years ago
Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?
V125BC [204]

Answer:

18 W

Explanation:

Applying,

P = V²/R.................. Equation 1

Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs

Since: It is a series circuit,

Then,

R = R1+R2............. Equation 2

Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

R = 8+8

R = 16 Ω

Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

P = 144/8

P = 18 W

7 0
2 years ago
What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2
kkurt [141]

Answer:

4.3\cdot 10^{-12} J

Explanation:

The fusion reaction in this problem is

4^1_1H \rightarrow ^4_2He +2e^+

The total energy released in the fusion reaction is given by

\Delta E = c^2 \Delta m

where

c=3.0\cdot 10^8 m/s is the speed of light

\Delta m is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

m(^1_1H)=1.007825u is the mass of one nucleus of hydrogen

m(^4_2 He)=4.002603u is the mass of one nucleus of helium

So the mass defect is:

\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u

The conversion factor between atomic mass units and kilograms is

1u=1.66054\cdot 10^{-27}kg

So the mass defect is

\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg

And so, the energy released is:

\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J

6 0
2 years ago
A 3.0-kilogram object is acted upon by an impulse having a magnitude of 15 newton•seconds. What is the magnitude of the object’s
Ilia_Sergeevich [38]
The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.
3 0
2 years ago
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