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2 years ago

A car is moving at a constant 34 m/s for 7 seconds. What is the average acceleration of the car?

Physics

1 answer:

Marina86 [1]2 years ago

7 0

Since the body is moving with uniform motion therefore the body doesn't accelerate
therefore acceleration is 0m/s^2

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how many times does the kinectic energy of a car increase when traveling 60 mph as opposed to traveling 30 mph

Bumek [7]

The car at 60 kph has 9 times more kinetic energy than the car traveling at 20 kph. This assumes that both cars have the same mass. Kinetic energy depends on the square of thee speed so if one car is going 3 times faster, its kinetic energy will be 3^2 ( = 9 ) greater. The car going at 60 kph will have 4 times the KE of the car going at 30 kph ( again assuming that the cars have the same mass.)

3 0

3 years ago

Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a

Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

$P = \frac{F}{A}$

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

$P = \frac{40\ N}{0.03\ m^2}\\\\$

4 0

3 years ago

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:

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