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Vikki [24]
2 years ago
13

What is the final velocity of a body if it is moving with 13 m/s in 300 seconds and its acceleration is 30 m/s 2.

Physics
1 answer:
Nutka1998 [239]2 years ago
6 0

Answer:

9013m/s

Explanation:

acceleration= v- u/ t

=> at = v-u

=> v = at + u

=> v =30*300+13

= 9013m/s

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To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

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R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Mila [183]

Answer:

v=6.65m/sec

Explanation:

From the Question we are told that:

Mass m=97.6

Coefficient of kinetic friction  \mu k=0.555

Generally the equation for Frictional force is mathematically given by

 F=\mu mg

 F=0.555*97.6*9.8

 F=531.388N

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 a_f=-\mu g

 a_f=-0.555 *9.81

 a_f=-54455m/sec^2

Therefore

 v=u-at

 v=0+5.45*1.22

 v=6.65m/sec

4 0
3 years ago
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