The strength of the gravitational field is given by:

where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:

And now we can use the previous equation to calculate the field strength at that altitude:

And we can see this value is a bit less than the gravitational strength at the surface, which is

.
Answer:
A) x4
Explanation:
Magnification is equal to image size divided by the actual size, or M = I/A.
The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:
28.8 cm/7.2 cm = 4
The area of the Earth (Ae) that is irradiated by is given by:
Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2
On the Earth, insolation (We) = Psun/Ae
Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W
Answer:
A = 2.36m/s
B = 3.71m/s²
C = 29.61m/s2
Explanation:
First, we convert the diameter of the ride from ft to m
10ft = 3m
Speed of the rider is the
v = circumference of the circle divided by time of rotation
v = [2π(D/2)]/T
v = [2π(3/2)]/4
v = 3π/4
v = 2.36m/s
Radial acceleration can also be found as a = v²/r
Where v = speed of the rider
r = radius of the ride
a = 2.36²/1.5
a = 3.71m/s²
If the time of revolution is halved, then radial acceleration is
A = 4π²R/T²
A = (4 * π² * 3)/2²
A = 118.44/4
A = 29.61m/s²
Answer:
Explanation: Determine the gravitational acceleration. ...
Decide whether the object has an initial velocity. ...
Choose how long the object is falling. ...
Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt