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yKpoI14uk [10]
3 years ago
7

A) Give 3 examples of forces that are pulls and 3 examples that are pushes. b) For each example you give, state an approximate v

alue for the size of the force.
Physics
1 answer:
Stels [109]3 years ago
7 0

Answer:

Explanation:

Force is defined as the push or pull which changes or tries to change the position, state of motion and the shape of the object.

(A) The examples of push are:

To push a chair on the floor, to push the car when it is stopped due to some problem, to push book on the table.

The examples of pull are :

To pull a chair towards you, to pull a string in a game of top, to pull the string in a gym.

(B) To push a chair or a book, the force required is small as compared to the to push a car.

To pull a chair or the string of top is less than the force to pull the string in gym.

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A
Pani-rosa [81]
The answer would be center of mass, B
8 0
3 years ago
Read 2 more answers
Ok plz answer and tell me how to do it
kirza4 [7]
Answer: 25N

method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.
4 0
3 years ago
The Demon Drop ride at the amusement park falls freely for 1.8 s after
timofeeve [1]

Answer:

Explanation:

a = -g = -9.80 m/s squared

d o  = 0

v o =  0

t = 1.8s

<h3>unkown:</h3>

d = ?

v = ?

8 0
3 years ago
Read 2 more answers
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
3 years ago
This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?
Crazy boy [7]
1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction

2) F=ma
= (5.5kg)(30m/s)
=165 N
7 0
2 years ago
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