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3 years ago

A person jogs at an average velocity of 3 m/s West. How long will it take this person to run a 30-meter city block?

Physics

2 answers:

grigory [225]3 years ago

8 0

Answer:

It will take the person 10 seconds to run a 30-meter city block.

Explanation:

To solve this problem, we must remember the formula for velocity:

velocity = displacement/time

If we plug in the values that we are given, we get:

3 m/s = 30 m/time

We can see from the above equation that our answer should be in seconds, so we can get rid of the units to simplify the equation slightly.

3 = 30/t

Next, we can multiply both sides by t:

3t = 30

Then, we should divide both sides of the equation by 3 to get the variable t alone on the left side of the equation.

t = 10

Therefore, the correct answer is 10 s.

Nataly_w [17]3 years ago

4 0

Answer:It will take a person 10 sec to run a 30-meter city block.

Explanation:

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3 years ago

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

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1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

$p_1=(2)(0)=0$

Using the same equation used in part 1), the initial momentum of the bowling ball is

$p=mv$

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

$p_2=(8)(20)=160 kg m/s$

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

$p_i = p_1 + p_2$

where

$p_1$ is the initial momentum of the trophy

$p_2$ is the initial momentum of the ball

Here we have

$p_1 = 0$

$p_2 = 160 kg m/s$

Therefore, the total momentum is

$p_i = 0 + 160 = 160 kg m/s$

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

$p_i = p_f$

where

$p_i$ is the total momentum before the collision

$p_f$ is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

$p_f = 160 kg m/s$