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sashaice [31]
2 years ago
12

motorcycle accelerates from rest to a final velocity of 27 m/s in 9 seconds. What is its acceleration?

Physics
1 answer:
Vedmedyk [2.9K]2 years ago
5 0

Answer:

3 m/s squared

Explanation:

The formula you use is Vf= Vi + at. You rearrange it to a= Vf - Vi/t. The Vf is 27m/s. The Vi is 0m/s and the t is 9s. Cross out Vi since it’s zero and you’re left with a= 27m/s divided by 9s, which equals 3

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One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee
musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

Speed = \frac{Distance}{Time}

\therefore 343 \ m/s= \frac{Distance}{5 \, s}

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is Time = \frac{Distance}{Speed} =  \frac{1715}{299,792,000} = 0.00000572 \, s which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0
3 years ago
2) Um gás ideal sofre uma determinada transformação, conforme mostra o gráfico abaixo. Considere
Roman55 [17]

Answer:

yes

Explanation:

5 0
3 years ago
Abcdefghijklmnopqrstuvwxyz
AlladinOne [14]

Answer:

lhgwljvqlivlajvliavpjavphvalhvqlhvqlhv

Explanation:

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6 0
3 years ago
Read 2 more answers
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
A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground.
Vaselesa [24]
V2=u2+2as
v2=144+600
v2=744
v=√744
6 0
3 years ago
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