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3 years ago

Quiz Revisions:

Physics

1 answer:

FrozenT [24]3 years ago

7 0

Answer:

fr = 514.5 N, This force has the opposite direction to the applied force.

Explanation:

Let's propose the solution of the problem using Newton's Second Law, we place a reference frame with the horizontal x-axis

Y axis

N- W = 0

N = W

X axis

F -fr = m a

the friction force has the expression

fr = μ N

fr = μ mg

fr = 0.35 150 9.8

fr = 514.5 N

This force has the opposite direction to the applied force.

The student's answer is incorrect because the friction coefficient must be multiplied by the normal

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Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 13.9 revolutions/s. Find the linear speed v

Lostsunrise [7]

Answer:

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Explanation:

Given;

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Question 30

stellarik [79]

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Explanation:

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$\delta=2 sin^{-1}(\frac{d}{2D})$

Where:

$d=16 m$ is the actual diameter

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Hence:

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3 years ago

The design of a 60.0 cm industrial turntable requires that it has a kinetic energy of 0.250 j when turning at 45.0 rpm. What mus

Aneli [31]

Answer:

The moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

Explanation:

Given;

radius of the turnable, r = 60 cm = 0.6 m

rotational kinetic energy, E = 0.25 J

angular speed of the turnable, ω = 45 rpm

The rotational kinetic energy is given as;

$E_{rot} = \frac{1}{2} I \omega ^2$

where;

I is the moment of inertia about the axis of rotation

ω is the angular speed in rad/s

$\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s$

$E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2$

Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

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3 years ago

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C. If the current is supplied by

Artemon [7]

Answer:

E = 20.03 J

Explanation:

Given that,

The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,

Voltage, V = 12 V

We need to find the energy delivered to the lightbulb filament during 2.00 s.

The energy delivered is given by :

$E=I^2Rt$ . ....(1)

As,

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As per Ohm's law, V = IR

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Using formula (1).

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So, the energy delivered to the lightbulb filament is 20.03 J.

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3 years ago