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2 years ago

How does the amount of friction affect the sum of forces

Physics

1 answer:

iren [92.7K]2 years ago

6 0

Answer:

<h3>Newton's 2nd law states acceleration is proportional to the net force acting on an object. The net force is the vector sum of all the forces applied to the object. ... In this case the acceleration (slowing down) of the puck is proportional to the amount of friction.</h3>

Explanation:

<h3>mark as brainliast</h3>

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What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 7400 N

bezimeni [28]

Answer: a = 1.32 * 10^18m/s² due north

Explanation: The magnitude of the force required to move the electron is given as

F = ma

The force exerted on the charge by the electric field of intensity (E) is given by

F = Eq

Thus

Eq = ma

a = E * q/ m

Where a = acceleration of charge

E = strength of electric field = 7400N/c

q = magnitude of electronic charge = 1.609 * 10^-6c

m = mass of an electronic charge = 9.109 * 10^-31kg

a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

a = 1.19 * 10^-12 / 9.019 * 10^-31

a = 0.132 * 10^19

a = 1.32 * 10^18m/s²

As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)

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The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in

MissTica

Answer:

Explanation:

Using the equation of motion to find the angular acceleration:

$\omega_f = \omega_i + \alpha t$

$\omega_f$ is the final angular velocity in rad/s

$\omega_i$ is the initial angular velocity in rad/s

$\alpha$ is the angular acceleration

t is the time taken

Given the following

$\omega_f = 6100rpm$

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x = 638.792rad/s

$\omega_f = 638.792rad/s\\$

Get the angular acceleration:

Recall that:

$\omega_f = \omega_i + \alpha t$

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>

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2 years ago

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