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3 years ago

Which statement describe. good hypothesis ? Check all that apply

1 answer:

7 0

A good hypothesis describes your ideas and how you think the experiment will conclude. Additionally, hypotheses have scientific information included.

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How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the

AnnyKZ [126]

Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.

7 0

3 years ago

Read 2 more answers

What can you observe to know that when clear methane gas is burned it is a chemical change?

aalyn [17]

<span>4) Formation of a gas

When methane burns, it produces the gases water vapor and carbon dioxide.
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6 0

3 years ago

If two or more elements are involved in a chemical reaction they will most likely form a?

Vitek1552 [10]

I know either mixture or compound

5 0

3 years ago

Read 2 more answers

Someone please help !!

masha68 [24]

I’m pretty sure it’s c.... hope it helps and hope it’s right.

7 0

3 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

3 years ago