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aleksklad [387]
4 years ago
8

In Paragraph 3 of this passage, what clues help you know the meaning of the word

Physics
1 answer:
n200080 [17]4 years ago
4 0
The answer is d because you have to make sure that everything is right
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The majority of women (64%) prefer trimmed pubic hair over natural or clean-shaven. ... Based on the results, it looks like a large majority of women prefer men to be performing some form of maintenance on their pubic hair region, but the adult film star clean-shaven look may not be the way to go.
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Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How far will the foot
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List two useful properties of current electricity and explain why.
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One example of current electricity are transmission lines. These bring electricity from power stations to individual houses.
6 0
4 years ago
HELP!! WILL MARK BRAINLIEST!!!!!
krek1111 [17]

Answer:

Create

Broken

Explanation:

Bond formation or creation requires the use of energy. Energy is used during bond formation between chemical species. The energy is required for the reaction to occur.

  • When bonds are broken, energy is released
  • Bond breaking process is a procedure that releases energy.
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5 0
3 years ago
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

3 0
3 years ago
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