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7nadin3 [17]
1 year ago
13

A student carries a very heavy backpack. To lift the backpack off the ground, the student must apply 80 N of force to do so. The

student lifts this backpack 1.5 m. How much work is done on the backpack?
Physics
1 answer:
Xelga [282]1 year ago
8 0

The work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.

<h3>How to calculate work done?</h3>

Work done is a measure of energy expended in moving an object; most commonly, force times distance.

It is said that no work is done if the object does not move, hence, the work done on an object can be calculated as follows:

Work done = Force × Distance

According to this question, a student carries a very heavy backpack and to lift the backpack off the ground, the student must apply 80 N of force to lift the backpack 1.5 m.

Work done = 80N × 1.5m

Work done = 120J

Therefore, the work done on the backpack by the student applies 80 N of force to lift the backpack 1.5 m is 120J.

Learn more about work done at: brainly.com/question/28172139

#SPJ1

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4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p
Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A ,   n 1  =  4 .2 mol

Number of moles mono atomic gas B ,   n 2  =  3.2mol

Initial energy of gas A ,   K A  =  9500  J

Thermal energy given by gas A to gas B ,   Δ K  =  600 J

Gas constant   R  = 8.314  J / molK

Let  K B  be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0
3 years ago
If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th
Trava [24]

Answer:

-  path differnce = 2.18*10^-6

-  1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

path\ difference\ = dsin\theta           (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m

- The maximum number of bright lines are calculated by using the following formula:

m\lambda = dsin\theta           (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769

The maxium number of bright lines are twice the previous result, that is, 1538 lines

8 0
3 years ago
Read 2 more answers
For a science project, you would like to horizontally suspend an 8.5 by 11 inch sheet of black paper in a vertical beam of light
liubo4ka [24]

Answer:

I = 3.9 x 10⁷ W/m²

Explanation:

given,

Sheet of black paper dimension = 8.5 x 11 inch

Area of sheet = 8.5 x 11 = 93.5 inch^2

1 inch =0.0254 m

Area = 0.06032 m²

mass of sheet = 0.80 g

Force = m g =  0.8 x 9.8 x 10⁻³ N

                    =  7.84 x 10⁻³ N

speed of light = c = 3 x 10⁸ m/s

Using equation

F = \dfrac{IA}{c}

where I is the intensity of light

7.84 \times 10^{-3} = \dfrac{I\times 0.06032}{3 \times 10^8}

2.352 \times 10^{6} = I\times 0.06032

I = \dfrac{2.352 \times 10^{6}}{0.06032}

I = 3.9 x 10⁷ W/m²

Intensity of the light is equal to I = 3.9 x 10⁷ W/m²

4 0
3 years ago
Which best explains why elements with a completely full outer shell of electrons do not usually bond with other elements to form
Alla [95]

The best answer is C.

The stability of atoms depends on whether or not their outermost shell is filled with electrons. If the outer shell is filled with electrons, the atom is stable and therefore they do not need to react  with other elements to become stable.

On the other hand, atoms with unfilled outer shells are unstable, and will usually form chemical bonds with other atoms to achieve stability. To achieve stability, atoms will form two types of chemical  bonds called ionic bonds and covalent bonds.





8 0
3 years ago
A planet outside of our solar system
Zolol [24]

Answer:

exoplanets is any planet beyond our solar system.

4 0
3 years ago
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