Answer:
300 Nm ; 300 J
Explanation:
Given that:
Force (F) = 20 N
Distance (d) = 15 m
The kinetic energy (Workdone) = Force * Distance
Kinetic Energy = 20N * 15m
Kinetic Energy = 300Nm
K. E = 1/2
Answer:
<h2>E) 52.5 cm</h2>
Explanation:
Step one:
given data
period T= 3 milliseconds= 0.003
velocity v= 175m/s
wave lenght λ=?
Step two:
we know that f=1/T
the expression relating period and wave lenght is
v=λ/T
λ=v*T
λ=175*0.002
λ=0.525m
to cm= 0.525*100
=52.5cm
The wavelength of the wave is E) 52.5 cm
Answer:
the renegade
Explanation: charklie dfamielo
First let us assign variables,
d = distance travelled
t = time it took
v = velocity of the commercial airline
In linear physics, the equation for velocity is given as:
v = d / t
Rewriting for d:
d = v t
We know that the distance to and from south America are equal
therefore:
d1 (going) = d2 (return)
Let us say that velocity of air is v3. Since going to South
America, the wind is against the direction of the plane and the return trip is
the opposite, therefore:
(v1 - v3) t1 = (v1 + v3) t2
(v1 – v3) 4 = (v1 + v3) 3.53
4 v1 – 4 v3 = 3.53 v1 + 3.53 v3
0.47 v1 = 7.53 v3
v1 = 16.02 v3
Since we also know that:
(v1 - v3) t1 = 784
(16.02 v3 – v3) * 4 = 784
60.085 v3 = 784
v3 = 13.05 mph
Therefore the speed of the plane in still air, v1 is:
v1 = 16.02 * 13.05
<span>v1 = 209.03 mph (ANSWER)</span>
<span> </span>
Answer: 
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is 
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity 
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
