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mojhsa [17]
3 years ago
7

A 4.88 x 10-6 C charge moves 265 m/s

Physics
2 answers:
rewona [7]3 years ago
7 0

Answer:

0 newtons

Explanation:

When a moving charge is parallel to the magnetic field, it feels no Magnetic force at all.

Even if the Magnetic Field is 100,000,000 Tesla!

Alex73 [517]3 years ago
5 0

Answer:

F=0N

Explanation:

From the question we are told that:

Charge Q=4.88 x 10-6 C

Velocity v= 265m/s

Angle \theta =0 \textdegree

Magnetic field B=0.0579T

Generally the equation for Force is mathematically given by

 F=Q(\=v*\=B)

 F=qvBsin\theta

Therefore

 F=qvBsin0 \textdegree

 F=0N

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Answer:

D

Explanation:

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Uma chapa metálica, com um furo central de diâmetro "d", é aquecida dentro de um forno. Com o aumento da temperatura, podemos af
Tatiana [17]

Answer:  O furo diminui enquanto a chapa aumenta a sua dimensão.

(The diameter decreases and the dimensions of the metal increases)

Explanation:

I will answer in English:

Here we have a piece of metal with a small hole on it, that is heated in a hoven.

We know that when a piece of metal is heated, it will expand (the density decreases, so the dimensions of the piece of metal increase).

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4 0
3 years ago
Person 1 moves an object 10 meters and exerts a force of 50N in 5 seconds. Person 2 moves an object 10 meters and exerts a force
mel-nik [20]

Answer:

Explanation:

Work is a force time the distance moved in the direction of that force, time is not a variable. Provided that the 50 N forces were applied in the same direction, the work done is identical. Assuming both applied force and direction of motion are horizontal W = Fd = 50(10) = 500J.

If the reason that one was slower is because the  second person applied his force at an angle, let's say 60° below the horizontal, then the work done by the second person is 50cos60(10) = 250 J

Time IS a consideration for Power, the RATE of doing work. Provided the force and motion are horizontal, the first person applied twice as much Power as the second person doing an identical amount of work in half the time.

7 0
3 years ago
A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla
kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v

Put the value into the formula

P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}

P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}

P=0.350\ Watt

Hence, The power output of the oscillator is 0.350 watt.

3 0
3 years ago
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