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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
Hey there friend :)
I would go with Choice A
Sorry if wrong :(
Have a Fabulous day!
Answer:
w'=(1/2)w
Explanation:
In order to calculate the angular velocity of the second wheel, you use the following formula:
(1)
v: speed of the wheel 1 = speed of the wheel 2
r: radius of the wheel 1
For the second wheel you have:
r'=2r
You replace this value of r' in the following equation:

The angular velocity of the second wheel is one half of the angular velocity of the first wheel
This question involves the concepts of centripetal force, range of projectile and projectile motion.
The magnitude of centripetal force is "2812.8 N".
First, we will find the velocity of the ball by using the formula of the range of the projectile.

where,
R = range of projectile = 86.75 m
v = speed = ?
θ = launch angle = 47.9°
g = acceleration due to gravity = 9.81 m/s²
Therefore,

v = 29.25 m/s
Now, we will use the formula to find out the centripetal force:

where,
= Centripetal Force = ?
m = mass of the ball = 7.3 kg
v = speed = 29.25 m/s
r = radius = 2.22 m
Therefore,

<u>Fc = 2812.8 N = 2.812 KN</u>
<u />
Learn more about centripetal force here:
brainly.com/question/11324711?referrer=searchResults
Answer:

Explanation:
Given
-- radius of hydrogen atom
<em>See attachment</em>
Required
Determine the radius of magnesium atom (R)
From the attachment, the ratio of a hydrogen atom to a magnesium atom is:

Simplify

Represent the radius as ratio:

Substitute 

Equate both ratios

Express as fraction

Cross Multiply







Hence, the radius of the magnesium atom is: 