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3 years ago

1. An 800-gram bowling ball is rolling in a straight line toward you. If its momentum is 16 kg•m/sec, how fast is it traveling?

Physics

1 answer:

ziro4ka [17]3 years ago

3 0

<u>Answer:</u>

The ball is rolling at a speed of 0.02 meter per second.

<u>Step by step explanation:</u>

We are given that there is a 800 gram bowling ball rolling in a straight line. If its momentum is given to be 16 kg.m/sec, we are to find its velocity.

For this, we will use the formula of momentum.

<em>Momentum = mass × velocity</em>

16 = 800 × velocity

Velocity = 16/800 = 0.02 meter per second

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denis-greek [22]

The center of mass isn't affected by the explosion.

To find the answer, we need to know about the trajectory of motion at zero external force.

<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>

When there's no net external force acting on an object, its momentum doesn't change with time.
As its momentum doesn't change, so it continues with the original trajectory.

<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>

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2 years ago

Calculate the impulse imparted when a 3,000-kg car hits a wall at 60 . m/s and comes to a stop.

Pani-rosa [81]

Impulse = change in momentum

The answer is 0.

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3 years ago

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the

ikadub [295]

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $\frac{velocity}{2 *length}$

velocity = $\sqrt{\frac{tension}{mass per unit length} }$

mass per unit length= $\frac{3.5}{1000*1.22}$ =0.00427 $\frac{kg}{m}$

Now calculating velocity v= $\sqrt{\frac{255}{0.00427} }$

=244.3 $\frac{m}{sec}$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $\frac{244.3}{2 *0.7}$ =174.5 hz

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3 years ago

A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much

nekit [7.7K]

Answer:

a $W_{3.5} = 490 \ J$

b $W_{2.5} = 250 \ J$

Explanation:

Generally the force constant is mathematically represented as

$k = \frac{F}{x}$

substituting values given in the question

=> $k = \frac{16}{0.2}$

=> $k = 80 \ N /m$

Generally the workdone in stretching the spring 3.5 m is mathematically represented as

$W_{3.5} = \frac{1}{2} * k * (3.5)^2$

=> $W_{3.5} = \frac{1}{2} * 80 * (3.5)^2$

=> $W_{3.5} = 490 \ J$

Generally the workdone in compressing the spring 2.5 m is mathematically represented as

$W_{2.5} = \frac{1}{2} * k * (2.5)^2$

=> $W_{2.5} = \frac{1}{2} * 80 * (2.5)^2$

=> $W_{2.5} = 250 \ J$

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3 years ago

A force of 2N will stretch a rubber band 0.02m. Assuming that Hooke's Law applies, answer the following: How far will a 1600N fo

denis-greek [22]

Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant

F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m

Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ

7 0

3 years ago