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3 years ago

When the direction of acceleration is opposite to the direction of velocity

Physics

2 answers:

wariber [46]3 years ago

7 0

Answer:

yes this is truly the wrong answer

nadezda [96]3 years ago

4 0

Answer:

If acceleration points in the same direction as the velocity, the object will be speeding up. And if the acceleration points in the opposite direction of the velocity, the object will be slowing down.

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Help for physical science u4 limiting reactants

Yuliya22 [10]

The reactants are on the left and the products are on the right of the equation

3 0

3 years ago

a car initially at 39.40 m/s accelerates at -3.00 m/s^2 how far has a car travel when it comes to a stop 9.35 seconds later

Kipish [7]

Answer:

499.523. meter

I hope it's helps you

6 0

3 years ago

What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at

romanna [79]

Answer:

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

6 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

2 years ago

Please help me with this question,giving the first correct answer brainliest

elena-s [515]

Answer:

Explanation:

4 0

2 years ago

Read 2 more answers