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Black_prince [1.1K]
3 years ago
12

You are pushing a heavy box across a rough floor. when you are initially pushing the box and it is accelerating, (a) you exert a

force on the box, but the box does not exert a force on you. (b) the box is so heavy it exerts a force on you, but you do not exert a force on the box. (c) the force you exert on the box is greater than the force of the box pushing back on you. (d) the force you exert on the box is equal to the force of the box pushing back on you. (e) the force that the box exerts on you is greater than the force you exert on the box.
Physics
2 answers:
kotegsom [21]3 years ago
4 0

Answer:

(c) the force you exert on the box is greater than the force of the box pushing back on you.

Explanation:

When a force is exerted on an object such that it starts accelerating without any change in the state of rest of the source of the force that always means that the reaction force of the object on the source is less than the action force.

This condition is in accordance with Newton's second law and Newton's third law.

kodGreya [7K]3 years ago
3 0

The answer is C. If the box is accelerating, that means that the amount of force you are exerting is greater than the force of the box.

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A teacher sends her students on a treasure hunt. She gives the following instructions:
Norma-Jean [14]

Answer:

D) Joe need to walk 481 m in a direction 40.9° north of east

Explanation:

Given:

1. Walk 300 m north

2. Walk 400 m northwest

3. Walk 700 m east-southeast and the treasure is buried there

Taking north as positive y axis and east as positive x axis.

Resolving their positions into x and y vector components.

1. d1 = 300j

2. d2 = -400cos45i + 400sin45j

3 d3 = 700cos22.5i - 700sin22.5j

Resultant position.

d = d1+d2+d3

d = (-400cos45 + 700cos22.5)i + (300+400sin45-700sin22.5)j

d = (363.873)i + (314.964)j

D = √( (363.873)^2 + (314.964)^2)

D =481.25m ~= 481m

Angle = taninverse (314.964/363.873)

Angle = 40.9°

Since the x and y components are both positive, it implies that their position is north of east

Therefore, Joe need to walk 481 m in a direction 40.9° north of east

7 0
3 years ago
A stone is thrown vertically upward with a speed of 24.0 ms. (a) How fast is it moving when it reaches a height of 13.0 m? (b) H
valina [46]

Answer:

Question A:17.92 m/s

Question B:0.6 seconds

Explanation:

Question A:

Initial velocity(u)=24m/s

Height(h)=13m

acceleration due to gravity(g)=9.8m/s^2

Final velocity=v

v^2=u^2-2xgxh

v^2=24^2-2x9.8x13

v^2=24x24-2x9.8x13

v^2=576-254.8

v^2=321.2

Take them square root of both sides

v=√(321.2)

v=17.92m/s

Question B:

velocity(v)=17.92m/s

Acceleration due to gravity(g)=9.8m/s^2

Initial velocity(u)=24m/s

Time=t

v=u-gxt

17.92=24-9.8xt

Collect like terms

9.8t=24-17.92

9.8t=6.08

Divide both sides by 9.8

9.8t/9.8=6.08/9.8

t=0.6 approximately

5 0
3 years ago
What happens to the piece of steel?
anyanavicka [17]
The answer should be c....
6 0
4 years ago
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Light with a wavelength of 494 nm in vacuo travels from vacuum to water. Find the wavelength of the light inside the water in nm
Mamont248 [21]

Answer:

370.6 nm

Explanation:

wavelength in vacuum = 494 nm

refractive index of water with respect to air = 1.333

Let the wavelength of light in water is λ.

The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.

By using the definition of refractive index

n = \frac{wavelength in air}{wavelength in water}

where, n be the refractive index of water with respect to air

By substituting the values, we get

1.333 = \frac{494}{\lambda }

λ = 370.6 nm

Thus, the wavelength of light in water is 370.6 nm.

8 0
3 years ago
What is the least value of an ammeter <br>Is it 1. 0.1<br>2. 0.10<br>3. 0.2<br>4. 0.5​
givi [52]

The least value varies depends on ammeter range. In the given question, ammeter range is not mentioned. So, the least value of an ammeter is 0.1 or 0.5 (depends on ammeter range).

<u>Explanation:</u>

The least value of an ammeter is the measure of the smallest number which can be observed in ammeter. So the least value will be varying depending upon its range. If we consider the range of ammeter is 30 A and the scale readings are 10 numbers, then the least value will be 30/10 = 3 A per scale.

So the least value is determined as \frac{Range of ammeter}{No.of scales}

So among the given options 0.1 is most suitable for an ammeter with range of 3 A with 30 divisions.

3 0
3 years ago
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