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Xelga [282]
3 years ago
7

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

. Assuming the acceleration is constant, how far did it travel during those 2.0 s?
Physics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

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3 years ago
A bus starts from rest.if the acceleration is 2m/s square, find
MrMuchimi

Answer:

The velocity after 2 seconds can be found through:

V = u +a*t

Where V is final velocity, u is initial velocity, a is acceleration and t is time.

V = 0 + 2* 2= 4 meters/second

The distance (s) can be found through:

V^2= u^2 +2*a* s

Where V is final velocity, u is initial velocity, a is acceleration.

4^2= 0^2 + 2 *2*s

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Distance (s) can also be found through:

s= ut + 1/2 at^2

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s= 4 meters

Explanation:

3 0
1 year ago
A 55-kg woman is wearing high heels.
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Answer:

Pressure, P=1.90\times 10^7\ Pa        

Explanation:

It is given that,

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Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{mg}{\pi r^2}

P=\dfrac{55\times 9.8}{\pi (0.003)^2}

P=1.90\times 10^7\ Pa

So, the pressure exerted on the floor is 1.90\times 10^7\ Pa. Hence, this is the required solution.

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Explanation:

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