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3 years ago

How do you reduce friction

1 answer:

3 0

To reduce friction between two surfaces, you have to
make the surfaces smoother.

The best way to do that is to introduce a fluid between them,
like grease, oil, or air.

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What do radio waves, microwaves, light, and x-rays have in common?

Dima020 [189]

One thing that radio waves, microwaves, light, and x-rays all have in common is that they are all forms of electromagnetic waves.

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4 years ago

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Block A, with a mass of 6.0 kg, is sliding across a frictionless track at 65 m/s when it collides with Block B ( 9.0 kg) which i

Mars2501 [29]

The correct answer is a I hope that helped enjoy the rest of your weekend

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3 years ago

Romi and Sony are two brothers. They wanted to go to the library. Just as they were

lora16 [44]

Answer:

Here are the answers you need

a. Romi's speed = slope of graph = 1000/15 - 0 = 66m/min

b. for 10 minute

c. I do not know the answer for this question, please forgive me

d.at 1000m from home they meet

I hope this helps you.

May god bless you and if you like this answer and think it helped you very much please mark me as brainliest because that will help me very much.

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3 years ago

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

NISA [10]

Answer:

The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If $\rho$ be the volume charge density,

Then, the charge will be,

$q=\rho\times\dfrac{4}{3}\pi R^3$ .....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

$\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}$

$E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}$

Put the value from equation (I)

$E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}$

$E=\dfrac{qr}{4\pi\epsilon_{0}R^3}$

Hence, The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

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3 years ago

Find u + (3v - w) u =-4i-8j v= -11i -3j w=5i+7j

k0ka [10]

Answer:

gunrun

Explanation:

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2 years ago