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elixir [45]
3 years ago
11

PLEASE HELP!!!!!!! A student is trying to demonstrate static electricity, so they rub two identical balloons with a neutral rabb

it fur to give them a charge. How
do these three objects now behave when brought near to each other?
A. They all attract each other.
B. The balloons repel each other but attract the rabbit fur.
O C. They all repel each other.
O D. The balloons attract each other but repel the rabbit fur.
Physics
1 answer:
vlada-n [284]3 years ago
4 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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The displacement z of a particle of rest mass m0, resulting from a constant force m0g along the z-axis is including relativistic
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Hi, your question was incomplete, hence I am writing down the complete question below.

Q)The displacement x of a particle of rest mass m0, resulting from a constant force m0g along the  x-axis, is

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<h3>Please refer to the attachment below.</h3>

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What is the law of conservation of energy?
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At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is
Katyanochek1 [597]

Answer:

4.47\ \text{km/h}

Explanation:

\dfrac{da}{dt} = Rate at which the distance between A and starting point of B is changing = -20 km/h

\dfrac{db}{dt} = Rate at which the distance of B is changing = 15 km/h

\dfrac{dc}{dt} = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = 20\times 4=80\ \text{km}

a = Distance remaining to the start point of B = 110-80=30\ \text{km}

b = Distance covered by B in 4 hours = 15\times 4=60\ \text{km}

Distance between A and B after 4 hours

c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}

c^2=a^2+b^2

Differentiating with respect to time we get

c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}

The rate at which the distance between the ships is changing at 4 PM is 4.47\ \text{km/h}.

7 0
3 years ago
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