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vodomira [7]
3 years ago
13

If I push on the wall with 75 Newtons of force, the wall will push back with __

Physics
1 answer:
Scilla [17]3 years ago
7 0

Answer: 0

Explanation:

75 newtons will push back canceling it out make it 0

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If you carry out an experiment measuring the weight and mass of objects in one particular location on the earth, what relation w
mario62 [17]
<span>Weight is directly proportional to mass.</span>
4 0
3 years ago
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What is wrong with the following statement: When you exert a force on a baseball, the equal and opposite force on the ball balan
gregori [183]

Answer:

When you exert a force on a baseball, there exists an equal and opposite force on the ball therefore, the ball will accelerate in opposite direction.

Explanation:

When you hit a ball with baseball bat, the bat exerts a great force on the ball which causes the ball to accelerate in the opposite direction. It is to be noted that the mass of bat is much greater than mass of ball but the acceleration of ball is also greater than the acceleration of the bat so both bat and ball almost exert same magnitude of force but in opposite direction and as a result both bat and ball accelerate in opposite direction, the deciding factor is of course the relative force applied by the batter and the bowler.

6 0
3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
I need help please it’s for a lab
krok68 [10]

Answer:

idk

Explanation:

5 0
3 years ago
The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x
Diano4ka-milaya [45]

Answer:

Ff=m\times \dfrac{V_o^2}{2X_1}

Explanation:

Given that

At X=0 V=Vo

At X=X1  V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

V^2=U^2-2aS

0=V_o^2-2a X_1

a=\dfrac{V_o^2}{2X_1}

So the friction force on the box

Ff= m x a

Ff=m\times \dfrac{V_o^2}{2X_1}

Where m is the mass of the box.

4 0
3 years ago
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