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Lemur [1.5K]
2 years ago
10

The wind blows a lawn chair that weighs 4 kg into a fence with a force of 8 N. How much reaction force does the fence exert on t

he chair?
Physics
1 answer:
AleksAgata [21]2 years ago
7 0

Answer:

40 N

explanation: force exerted on chair = mass of chair × acceleration

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9. How much work is done when a 15kg box is lifted to a height of 2 meters?
torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd        where F = mg

W= Fd

   = mg d

   = 15 kg ( 9.8 m/s²) ( 2m )

   = 294 J

5 0
3 years ago
A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.
jenyasd209 [6]
<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

  • Force of the dog is 24 N
  • Distance upward is 4.9 m

We are required to calculate the work done

  • Work done is the product of force and distance
  • That is; Work done = Force × distance
  • It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

                  = 117.6 Joules

Hence, the work done in lifting the dog is 117.6 Joules

3 0
3 years ago
THIS IS MY EXAM HURRY PLS
tangare [24]

Answer:

elements in the same column have the same number of neutrons. elements with similar mass are placed in the same column.

5 0
2 years ago
Read 2 more answers
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
2 years ago
As an object falls:
alex41 [277]

Answer:

c it is not accelerating on it's on but gravity pulls it there for velocity increases.

5 0
2 years ago
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