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pentagon [3]
4 years ago
13

An object at rest has no net force acting on it. True False

Physics
2 answers:
enot [183]4 years ago
6 0

Answer:

True

Explanation:

Unless acted by another force a body moving at constant velocity or at rest remains at that state. So, yes an object at rest has no net force acting on it. Only when another force is acted on the object the object will move. This property is known as inertia. When the body is at rest it is known as inertia of rest and inertia motion when it is moving.

aivan3 [116]4 years ago
5 0
Yes that's true according to newton's first law 
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adoni [48]
Her speed was 7.27 meters per second
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3 years ago
Могут ли быть сообщающиеся сосуды неодинаковы пр своей форме
Rudik [331]

Answer:

no

Explanation:

they cannot because they contain the same amount of liquid

8 0
3 years ago
Read 2 more answers
If Fg=mg solve for g
Bas_tet [7]

Answer:Fg = mg however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that Fg = mg because certainly not for every single situation the net force is going to equal to the force of gravity please explain... what allows us to say Fg = mg

Source https://www.physicsforums.com/threads/fg-mg-questioned.336776/

Explanation:

6 0
3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
3 years ago
If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later
garik1379 [7]
X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

5 0
3 years ago
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