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3 years ago

Is it correct to say that constant speed = 0 acceleration = no resultant force?

Physics

1 answer:

stepan [7]3 years ago

8 0

Nope......... Constant speed but change in direction can cause acceleration which would give a finite force........

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4. What is the momentum of a 70 kg object traveling at 20 m/s?

Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

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A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

dangina [55]

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

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And m2 and v2 be the mass and velocity of the bowling pin

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2 years ago

If two objects have the same mass, what determines the strength of the gravitational force between them? (4 points) Select one:

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2 years ago

If you carry out an experiment measuring the weight and mass of objects in one particular location on the earth, what relation w

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3 years ago

Read 2 more answers

A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi

Dmitriy789 [7]

Answer:

$A = 0.081 \ m$

The value is $u = 0.2569 \ m/s$

Explanation:

From the question we are told that

The mass is $m = 0.750 \ kg$

The spring constant is $k = 17.5 \ N/m$

The instantaneous speed is $v = 39.0 \ cm/s= 0.39 \ m/s$

The position consider is x = 0.750A meters from equilibrium point

Generally from the law of energy conservation we have that

The kinetic energy induced by the hammer = The energy stored in the spring

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2$

Here a is the amplitude of the subsequent oscillations

=> $A = \sqrt{\frac{m * v^ 2 }{ k} }$

=> $A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }$

=> $A = 0.081 \ m$

Generally from the law of energy conservation we have that

The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2$

=> $\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2$

=> $u = 0.2569 \ m/s$

3 0

2 years ago