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zzz [600]
3 years ago
8

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car's motion, the east

ward component of the car's velocity is given by υx(t)=(0.980m/s3)t2.
What is the acceleration of the car when υx = 14.5 m/s ?
Physics
1 answer:
wel3 years ago
7 0

Answer:

7.54 m/s²

Explanation:

uₓ(t) = (0.980 m/s³) t²

Acceleration is the derivative of velocity with respect to time.

aₓ(t) = 2 (0.980 m/s³) t

aₓ(t) = (1.96 m/s³) t

When uₓ = 14.5 m/s, the time is:

14.5 m/s = (0.980 m/s³) t²

t = 3.85 s

Plugging into acceleration equation:

aₓ = (1.96 m/s³) (3.85 s)

aₓ = 7.54 m/s²

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                          I = Current ; in Amperes

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4 years ago
The Surface Pressure at Leh, Ladakh is 800 mb. Now, assuming that Leh is at an altitude of 3500 m and every 100 m increase in he
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We have that the sea level pressure for Leh area is 1150mb mathematically given as

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<h3> Sea level pressure</h3>

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Ladakh is 800 mb.

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increase in height with respect to sea level corresponds to 10 mb pressure,

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Therefore,  here for the sea level <em>pressure</em> we need to add,

Ps=800+350

Ps= 1150 mb

For more information on Pressure visit

brainly.com/question/25688500

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