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Oksana_A [137]
3 years ago
10

An object originally at rest, is accelerated uniformly along a straight line to a speed of 8m/s in 2s. What is the acceleration

of the object?​
Physics
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

4m/s²

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 8 m/s

Time taken (t) = 2 sec

Acceleration (a) = ?

We know

a  =  \frac{v - u}{t}  \\  =  \frac{8 - 0}{2} \\  =  \frac{8}{2}  \\  = 4 \: m |s ^{2}

Hope it will help :)

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-- A moving electric field creates a magnetic field.

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An AED should only be used if a person is too tired to continue with chest compressions.
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The given statement "An AED should only be used if a person is too tired to continue with chest compressions" is false.

Answer: Option B

<u>Explanation:</u>

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4 years ago
Playing soccer on a beach will require more effort because the sand causes a great deal of _______ to the ball.
34kurt

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I believe the answer to be B

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If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry

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2.486 L is equal to:
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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
aksik [14]

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y =  \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} }  =2\ m/s

Also:

\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x =  u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s

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