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gregori [183]
2 years ago
14

The _____ of an object consists of its speed and direction.

Physics
1 answer:
ELEN [110]2 years ago
3 0
Velocity would be an answer
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According to Ohm's law, a circuit with a high resistance _____. will have a low electric current will have a high electric curre
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Remark
When you are asked a question like this, the first thing to do is search out a formula and put some limits on it.

Formula
I = E/R which comes from E = IR. To get to the derived formula, divide both sides by R
E/R = I*R/R
E/R = I

Discussion
This is an inverse relationship. That means that as one goes up the other one will go down. 

So in this case you keep E constant and you manipulate R and look at your results for I

Case 1

Let us say that E = 10 volts
Let us also say the R = 10 ohms

I = E/R
I = 10/10
I = 1 ohm

Case Two
Let's raise the Resistance to 100 ohms
E = 10
R = 100
I = 10/100 = 0.1 

Conclusion
As the Resistance goes up, the current goes down. Answer: A
8 0
3 years ago
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Trumpeter A holds a B-flat note on the trumpet for a long time. Person C is running towards the trumpeter at a constant velocity
Vikki [24]
You didn't mention it, but the trumpeter herself has to be standing still.

<span>Person C, the one running towards the trumpeter, hears a pitch
that is higher than B-flat.  (A)

Person B, the one running away from the trumpeter, hears a pitch
that is lower than B-flat.

Person D, the one standing still the whole time, hears the B-flat.</span>
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Sally and Juan’s science teacher has directed the class to build a paper airplane. The challenge is to build an airplane that fl
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During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25
Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

dQ=0

dU=-dW

We need to calculate the number of moles and specific heat

Using formula of heat

dU=nC_{v}dT

nC_{v}=\dfrac{dU}{dT}

Put the value into the formula

nC_{v}=\dfrac{-100}{5}

nC_{v}=-20\ J/K

We need to calculate the heat

Using formula of heat

dQ=nC_{v}(dT_{1})+dW_{1}

Put the value into the formula

dQ=-20\times5+25

dQ=-75\ J

We need to calculate the heat capacity for the second process

Using formula of heat

dQ=nC_{v}(dT_{1})

Put the value into the formula

-75=nC_{v}\times(-5)

nC_{v}=\dfrac{-75}{-5}

nC_{v}=15\ J/K

Hence, The heat capacity for the second process is 15 J/K.

5 0
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