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3 years ago

7

A ball of radius 0.220 m rolls along a horizontal table top with a constant linear speed of 3.70 m/s.?

1 answer:

4vir4ik [10]3 years ago

6 0

<span>16.82 x 0.04 = 0.67 rad
I hope I helped if you really need I can explain to you how I got that answer but Thats correct im sorry it took 2 days for me to find this answer but if you or anybody else still needs the answer for this question here it is :) have a fantastic day guys Spring Break is coming up soon :)</span>

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How much does coast to coast membership cost?

CaHeK987 [17]

The price of coast to coast membership in united states could lie anywhere between $2,000 to $ 5,000
Unless you're a frequent user of this type of event, i think it would be economically more efficient if you pay the resort on one-day price

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3 years ago

PLEASE HELP While the Earth is revolving around the sun, less direct sunlight is reaching the Northern Hemisphere than the South

jekas [21]

Answer:

Winter

Explanation:

Earth Rotates about an axis

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3 years ago

He power output of a cyclist moving at a constant speed of 5.3 m/s on a level road is 120 w. (a) what is the force exerted on th

EleoNora [17]

P=F*V
F=P/V
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8 0

3 years ago

4. What is the electric field strength 1.4 nm from a charge of 4.7 cC?

pentagon [3]

The electric field strength is $2.16\cdot 10^{26} N/C$

Explanation:

The strength of the electric field produced by a single point charge is given by:

$E=k\frac{q}{r^2}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge at which the field strength is calculated

For the charge in the problem, we have:

$q=4.7 cC = 0.047 C$ is the charge

$r=1.4 nm = 1.4\cdot 10^{-9} m$

Therefore, the electric field strength is

$E=(8.99\cdot 10^9)\frac{0.047}{(1.4\cdot 10^{-9})^2}=2.16\cdot 10^{26} N/C$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

Suppose that light from a laser with wavelength 633 nm is incident on a thin slit of width 0.500 mm. If the diffracted light pro

coldgirl [10]

Explanation:

It is given that,

Wavelength, $\lambda=633\ nm=633\times 10^{-9}\ m$

Slit width, $a=0.5\ mm=0.0005\ m$

Order, m = 2

If the diffracted light projects onto a screen at distance 1.50 m, L = 1.5 m

For the diffraction of light,

$y=\dfrac{m\lambda L}{a}$

$y=\dfrac{2\times 633\times 10^{-9}\times 1.5}{0.0005}$

y = 0.0037 m

So, the distance from the center of the diffraction pattern to the dark band is 0.0037 meters. Hence, this is the required solution.

3 0

3 years ago