Given Information:
Elapsed time = t = 6 seconds
Required Information:
Distance = d = ?
Answer:
Distance = d = 2.058 km
Explanation:
We know that the speed of sound in the air is given by
v = 343 m/s
The relation between distance, speed and time is given by
distance = speed*time
substituting the given values yields,
distance = 343*6
distance = 2058 m
There are 1000 meters in 1 km so
d = 2058/1000
d = 2.058 km
Therefore, the storm is about 2.058 km away when elapse time between the lightning and the thunderclap is 6 seconds.
(AB) & (CD)
Most machines are fueled by gasoline and electricity
<span>373.2 km
The formula for velocity at any point within an orbit is
v = sqrt(mu(2/r - 1/a))
where
v = velocity
mu = standard gravitational parameter (GM)
r = radius satellite currently at
a = semi-major axis
Since the orbit is assumed to be circular, the equation is simplified to
v = sqrt(mu/r)
The value of mu for earth is
3.986004419 Ă— 10^14 m^3/s^2
Now we need to figure out how many seconds one orbit of the space station takes. So
86400 / 15.65 = 5520.767 seconds
And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity
2 pi r / 5520.767
Finally, combining all that gets us the following equality
v = 2 pi r / 5520.767
v = sqrt(mu/r)
mu = 3.986004419 Ă— 10^14 m^3/s^2
2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r)
Square both sides
1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r
Multiply both sides by r
1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2
Divide both sides by 1.29527 * 10^-6 s^2
r^3 = 3.0773498781296 * 10^20 m^3
Take the cube root of both sides
r = 6751375.945 m
Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So
6751375.945 m - 6378137.0 m = 373238.945 m
Converting to kilometers and rounding to 4 significant figures gives
373.2 km</span>
K.E = 1/2 mv²
800 = 1/2 ×12 ×v²
800 = 6 v²
800 / 6 = v²
= 133.4 =v²
√133.4 = √v²
11.5 = v²
I hope this answer is correct.
Answer:
<em>Correct choice: b 4H</em>
Explanation:
<u>Conservation of the mechanical energy</u>
The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):
E = U + K
The GPE is calculated as:
U = mgh
And the kinetic energy is:
Where:
m = mass of the object
g = gravitational acceleration
h = height of the object
v = speed at which the object moves
When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:
When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:
Since the energy is conserved, U1=U2
![\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%20%20%20%5Cqquad%5Cqquad%20%5B1%5D)
For the speed to be double, we need to drop the snowball from a height H', and:
Operating:
![\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%27%3D4%5Cfrac%7B1%7D%7B2%7Dm%28v%29%5E2%20%5Cqquad%5Cqquad%20%5B2%5D)
Dividing [2] by [1]
Simplifying:
Thus:
H' = 4H
Correct choice: b 4H
