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3 years ago

6

For every action there is______and opposite reaction

1 answer:

Ymorist [56]3 years ago

5 0

for every action there is an equal and opposite reaction

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Doubling the distance between you and a source of radiation decreases your exposure by:

Elan Coil [88]

Given what we know, we can confirm that doubling the distance between you and a source of radiation decreases your exposure by 75%.

<h3>How is distance related to radiation exposure?</h3>

As expected, increasing the distance from the source of the radiation will reduce its negative effects.
Counter-intuitively however, doubling the distance does not reduce by half, but rather reduces its effects by 3/4th.
This is due to the fact that the radiation effects from the source are inversely proportional to the square of the distance.
This causes the changes to be far greater than expected.

Therefore, given that the radiation is proportional to the square of the distance, instead of being of a more direct relation, we can confirm that when doubling the distance between yourself and the source of the radiation, you can reduce its effects by 3/4 or 75%.

To learn more about radiation visit:

brainly.com/question/9815840?referrer=searchResults

5 0

3 years ago

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

3 0

3 years ago

State three precautions taken when using a metre rule

marishachu [46]

Answer:

1. Our eye should be vertically abvove the point, where the measurement is to be taken.

2. For proper measurement, Keep the ruler exactly along the length , for the measurement

Explanation:

good luck

5 0

3 years ago

A boat heads directly eastward across a pier at 12 meters per second. If the current in the river is flowing

Trava [24]

Answer: 13 m/s

Explanation: The two vectors form a 5-12-13 right triangle. the magnitude of their resultant is the hypotenuse, which is 13 m/s.

4 0

4 years ago

A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.

Anettt [7]

Answer:

The average acceleration is 16.6 m/s² ⇒ 1st answer

Explanation:

A rocket achieves a lift-off velocity of 500.0 m/s from rest in

30.0 seconds

The given is:

→ The initial velocity = 0

→ The final velocity = 500 meters per seconds

→ The time is 30 seconds

Acceleration is the rate of change of velocity of the rocket

→ $a=\frac{v-u}{t}$

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

→ u = 0 , v = 500 m/s , t = 30 s

Substitute these values in the rule

→ $a=\frac{500-0}{30}=\frac{500}{30}=16.6$ m/s²

<em>The average acceleration is 16.6 m/s²</em>

6 0

3 years ago