The moons volume is that of 2 percent of the earth.
Answer:
a.After 
second Mr Comer's speed 
b.Distance travelled by Mr.Comer in seconds 
Explanation:
a. Lets recall our first equation of motion 
Now we know that 
, 
and
Plugging the values we have.
Then Mr.Comer's speed after sec 
b.
Lets find the distance and recall our third equation of motion.
So 
distance covered.
Dividing both sides with 2a we have.
Plugging the values.
So Mr.Comer will travel a distance of .
Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels <em>x</em> after time <em>t</em> horizontally is
<em>x</em> = (7.2 m/s) <em>t</em>
The object's height <em>y</em> at time <em>t</em> is
<em>y</em> = 9.4 m - 1/2 <em>gt</em>²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is
<em>v</em> = -<em>gt</em>
(a) The object hits the ground when <em>y</em> = 0:
0 = 9.4 m - 1/2 <em>gt</em>²
<em>t</em>² = 2 * (9.4 m) / (9.80 m/s²)
<em>t</em> ≈ 1.92 s
at which time the object's vertical velocity is
<em>v</em> = -<em>g</em> (1.92 s) = -18.8 m/s ≈ -19 m/s
(b) See part (a); it takes the object about 1.9 s to reach the ground.
(c) The object travels a horizontal distance of
<em>x</em> = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m
For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
Mercury and Venus are therefore closer to each other most of the time. But Earth is the planet closest to Venus. And that's why from here on Earth, Venus looks so big and luminous. Venus is the brightest thing in the night sky after the sun and the moon.
