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3 years ago

If an object is not moving are the forces acting on it balanced? Yes or no?why?

Physics

1 answer:

xxMikexx [17]3 years ago

6 0

This is another time to look at Newton's 2nd law of motion:

Net Force = (mass) x (acceleration)

If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:

Net Force = (mass) x (zero)

or Net Force = (zero) .

"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.

So the answer is '<em>yes</em>', and that's why.

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A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca

Mariulka [41]

i believe it is C....tell me if im right please<3

4 0

3 years ago

Read 3 more answers

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca

Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

$a=\frac{-\mu mg}{m}=-\mu g$

where:

$\mu=0.80$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting and solving for s, we find:

$s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m$

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

$\mu=0.25$

Here we want the stopping distance of the car to remain the same as part a), so

$s=52.5 m$

We can use again the same suvat equation:

$v^2-u^2=2as$

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

$v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s$

7 0

3 years ago

PLEASE ANSWER THIS ASAP!!!

Burka [1]

Answer:

I guess B

Explanation:

Because it stop by it self so that is digital

5 0

2 years ago

Elements in group ____ are known as salt formers

Zinaida [17]

The group that is known as a salt former is the halogens group

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3 years ago

Katie and Mark sit next to one another in class. She has a mass of 40 kg and his mass is 65 kg.

OlgaM077 [116]

A :-) for this question , we should apply
F = GMm by d^2
( For making the calculation easy , first remove the decimals )
Given : G = 6.7 x 10^-11 Nm^2 / kg^2
= 67 x 10^-12 Nm^2 / Kg^2
M = 65 kg
m = 40 kg
d = 0.5 m
Solution -
F = GMm by d^2
F = 67 x 10^-12 x 65 x 40 by 0.5 x 0.5
F = 4355 x 40 x 10^-12 by 0.25
F = 174200 x 10^-12 by 0.25
F = 696800 x 10^-12

.:. The Gravitational force between mark and Katie is 696800 x 10^-12

6 0

3 years ago