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8 months ago

Two positive charge spheres, the spheres are separated by 0.40 m. The charge on the first sphere is 100 microcoulombs and the

Physics

1 answer:

agasfer [191]8 months ago

8 0

Answer: 168.75 N

Explanation:

first, let's convert microcoulombs to coulombs

q1 = 1e-4 C

q2 = 3e-5 C

r = 0.4 m

then use the equation Fe = $\frac{kq_{1} q_{2}}{r^{2} }$

plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2

F = 168.75 N

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A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of

Neko [114]

The answer is letter C. A train travels from the bottom of a hill to the top of a hill, its moves slower by the time it reaches the top is an example of negative acceleration.

In physics, acceleration can be described as an objects change of velocity. When an object gains velocity, it is positive acceleration, and negative acceleration for the opposite.

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2 years ago

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

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time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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3 years ago

If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average

Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

$a=\frac{V-V0}{t}$

V=0 m/s (because the frog stopped)

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So;

$a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V$

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2 years ago

A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is

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The period will be the same if the amplitude of the motion is increased to 2a

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Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

$T = 2\pi \sqrt{\frac{m}{k} }$

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

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Increasing the amplitude does not affect the period of the mass and spring system.

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2 years ago

How much time will elapse between seeing and hearing an event?

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Depends on how far away the event is and what the temperature is as this affects the speed of sound.

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3 years ago