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vivado [14]
3 years ago
13

A car moving eastward along a straight road increases its speed uniformly from 16 m/s to 32 m/s in 10.0 s.

Physics
1 answer:
kirill115 [55]3 years ago
7 0

Answer:

a)a= 1.6 m/s^2

b)V= 24 m/s

c)d=240 m

Explanation:

Given that

u = 16 m/s

v= 32 m/s

t= 10 s

The average acceleration = a

We know v = u +at

32 = 16 + a x 10

a=\dfrac{32-16}{10}\ m/s^2

a= 1.6 m/s^2

d=ut+\dfrac{1}{2}at^2

d=displacement

d=16\times 10+\dfrac{1}{2}\times 1.6\times 10^2

d=240 m

The average velocity

V=\dfrac{d}{t}

V=\dfrac{240}{10}\ m/s

V= 24 m/s

a)a= 1.6 m/s^2

b)V= 24 m/s

c)d=240 m

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Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given \Delta x=7\ m

\theta=45 \°

Also, \Delta y=(3.5-2)=1.5\ m

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\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s.

Plugging this value in

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t=1.06\ seconds

So, the time of flight of the ball is 1.06 seconds.

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