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SVEN [57.7K]
3 years ago
8

At one point in the circuit, there is an LED and a resistor in parallel with one another. If you measure the voltage drop across

the LED as 2.1 V and the current through the LED as 20 mA, you know: Group of answer choices a) The current through the resistor is 20 mA b) The voltage drop across the resistor is 2.1 V and its current is 20 mA c) The voltage drop across the resistor is 2.1 V nothing about the current through or voltage drop across the resistor.
Physics
1 answer:
kifflom [539]3 years ago
7 0

Answer:

C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.

Explanation:

When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.

If it were a series connection, then the current would have been the same, but the voltage drop were another story.

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A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo
viva [34]

The fundamental frequency on a vibrating string is given by:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension

\mu is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor \sqrt{2}

In this case, the mass per unit length is doubled:

\mu'=2\mu

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}

So, the fundamental frequency will decrease by a factor \sqrt{2}.

(c) the fundamental frequency will increase by a factor \sqrt{2}

In this case, the tension is doubled:

T'=2T

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f

So, the fundamental frequency will increase by a factor \sqrt{2}.

8 0
3 years ago
A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.
vekshin1

Answer:

A)  If the paintball stops completely the magnitude of the change in the paintball’s momentum is  p=0.273kg*m/s

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is  p=0.546kg*m/s

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

Hi

A) We use the formula of momentum p=mv, so we have p=0.0032kg*85.3m/s=0.273kg*m/s

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

p=2*0.0032kg*85.3m/s=0.546kg*m/s.

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

F=\frac{\Delta p}{\Delta t}, as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

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4 years ago
1. You and your forensics team are working a crime scene of a 19 year-old female who was found in the woods behind
Helga [31]

Answer:

You could conclude the 2nd fact

Explanation:

3 0
3 years ago
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