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ipn [44]
3 years ago
11

An object is traveling at a constant velocity of 12 m/s when it experiences a constant acceleration of 2.5 m/s 2for a time of 20

s. What will its velocity be after that acceleration? SHOW WORK. ONLY SERIOUS RESPONSES. DUMB COMMENTS WILL BE DELETED. THINGS THAT ARE N/A WILL BE DELETED.
Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

62m/s

Explanation:

If the object accelerates at a constant rate of 2.5 m/s^2 for 20 seconds, then it will speed up by a total of 2.5*20=50m/s. Adding this to the initial velocity of 12m/s, you get a total velocity of 62m/s. Hope this helps!

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On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou
Feliz [49]

Answer:

Angular velocity, N_f = 242.36 rpm

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, m_{a} = 0.13 * \frac{M}{2} ( since it is 13% of the whole body and each arm is considered)

m_{a} = 0.13 * 37\\m_a = 4.81 kg

Mass of the trunk, m_{t} = M - 2m_{a}

m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2

(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2

The final moment of inertia of the person:

(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2

According to the principle of conservation of angular momentum:

(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm

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2 years ago
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