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Monica [59]
3 years ago
6

HOW DOES ALCOHOL AND DRUGS PLAY A FACTOR IN SEXUAL ASSAULT

Physics
1 answer:
Naily [24]3 years ago
5 0

Answer:

Alcohol and drugs play a factor in sexual assault through bad ideas without education (such as committing the assault), as well as not being conscious when you are a victim of sexual assault.

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Explanation:

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3 years ago
Which of the following is an example of energy being transferred by sound waves? A loud clap of thunder causes a window to vibra
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All of them are examples of energy being transferred by waves<span />
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As the rate of radioactive decay becomes smaller, half-life becomes ____.
Marina86 [1]

Answer:

Larger

Explanation:

8 0
3 years ago
A particle having a speed of 0.87c has a momentum of 10-16 kg·m/s. what is its mass?
cupoosta [38]
The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
p=mv
In our problem, we know p=10^{-16}~Kg \cdot m/s and v=0.87c=0.87\cdot 3\cdot10^8~m/s=2.61\cdot 10^8~m/s, and using the relationship mentioned above, we can find the mass m of the particle:
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8 0
3 years ago
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} }

7 0
3 years ago
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