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FrozenT [24]
3 years ago
7

Which of the following is typically NOT a method scientist use to determine astronomic distances

Physics
1 answer:
tangare [24]3 years ago
5 0
I think the answer is A
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A sound wave increases in pitch. Which of the following changes has occurred?
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How do objects with the same charge interact? How do objects with opposite charges interact?
Marta_Voda [28]
Objects with the same charge repels each other whereas objects with opposite charges attract each other.

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A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s, how high is the cliff, if it took 12 s for the
Vesna [10]

Answer:

The height of the cliff from which the ball was dropped from is 224.4m.

\overline{v}={\frac{\Delta x}{\Delta t}}

Given the data in the question;

Initial velocity of the ball;

Time taken by the ball to reach the ground;

Distance or Height of the cliff from which the ball was thrown from;

To get the height of the Cliff, we use the Second Equation of Motion:

Where s is the distance or height,  is the initial velocity, t is the time and a is the acceleration. Since the ball was thrown down from a certain height (cliff), its is now under the influence of gravity. acceleration due to gravity;

Hence, the equation becomes

We substitute the given values into the equation

Therefore, the height of the cliff from which the ball was dropped from is 224.4m

Explanation:

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2 years ago
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An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

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2 years ago
If a fuse melts, does it create an open circuit, a closed circuit, or a short circuit?
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Answer:

short circuit

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