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Scorpion4ik [409]

2 years ago

6

Please select the word from the list that best fits the definition

2 answers:

vagabundo [1.1K]2 years ago

7 0

Answer:input work

Explanation:

Diano4ka-milaya [45]2 years ago

3 0

Answer:

thankssssssssss\

Explanation:

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Am I correct?? Will give brainliest

IRISSAK [1]

You are correct earth science is studied to predict planetery changes

4 0

2 years ago

Read 2 more answers

I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow

Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0

3 years ago

Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.

Firlakuza [10]

Answer:

$s = 414.7 m\\$

Explanation:

The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:

s = rθ

where,

s = distance traveled on road = ?

r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m

θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad

Therefore,

$s = (1.1 m)(377 rad)\\s = 414.7 m$

8 0

2 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

2 years ago

Can you please help me figure out the work for the missing quantities?

marysya [2.9K]

Answer: set up proportions

Explanation:

8 0

2 years ago