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2 years ago

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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damaskus [11]

Answer:

$a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

Explanation:

According to Newton's Third Equation of motion

$V_f^2-V_i^2=2as$

Where:

$V_f$ is the final velocity

$V_i$ is the initial velocity

a is the acceleration

s is the distance

In our case:

$V_f=V_{escape}, V_i=0,s=4 m$

So Equation will become:

$V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

5 0

3 years ago

In order to identify them, animal fossils and other fossils nearby can be: A. compared B. piled C. seen

tigry1 [53]

Answer:

A.compared

Explanation:

Fossils help figure out the time that organisms lived. If you know one of the fossils, it can be used as a reference for others around.

5 0

3 years ago

HELP!!! Why is plasma considered the most common state of matter in the universe?

olasank [31]

Clue : There is plasma in sun

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3 years ago

Read 2 more answers

A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$