According to the Law of Universal Gravitation, the gravitational force is directly proportional to the mass, and inversely proportional to the distance. In this problem, let's assume the celestial bodies to be restricted to the planets and the Sun. Since the distance is specified, the other factor would be the mass. Among all the celestial bodies, the Sun is the most massive. So, the Sun would cause the strongest gravitational pull to the satellite.
Answer:
500 N
Explanation:
Given that,
The upward force is 800 N and the downward forces are 400 N, 500 N, 400 N.
At equilibrium, the upward forces will become equal to the downward forces. Let the reading in the right hand scale.
x + 800 = 400 + 500 + 400
x + 800 = 1300
x = 1300 - 800
= 500 N
So, the reading in the right hand scale is 500 N.
Answer:
The average velocity is
and respectively.
Explanation:
Let's start writing the vertical position equation :
Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt =
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :
For the position variation we use the vertical position equation :
Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is
For the second time interval :
t1 = 4 s → t2 = 9 s
Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :
Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then
The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
The answer to your question Tropical Cyclones
Answer:
1.27 Mega watt approx
Explanation:
Given data
Mass= 0.5kg
c= 4200 J/kg °C
T1= 100°C
Time = 2 hours >>>> 7200 seconds
From the given data the water is already at the boiling point
Hence if continuous heat is being added the latent heat of vapourization will result, the latent heat of vaporization is about 2,260 kJ/kg
Q= mL
Q= 0.5*2,260
Q = 1130kJ
Power= Energy * Time
Power= 1130* 1130
Power= 1.27mega watt approx