<span>b) The force with a distance of 150 km is 889 N
c) The force with a distance of 50 km is 8000 N
This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question.
Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as
F = (G M1 M2)/r^2
where
F = Force
G = gravitational constant
M1 = Mass 1
M2 = Mass 2
r = distance between center of masses for the two masses.
So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889.
Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits.
If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>
Answer:
29.412m/s
Explanation:
where F= force, m= mass, and a=acceleration
we also know that,
a = Δv / t where Δv = change in velocity and t = time
thus F = m ( Δv / t)
Δv
29.412m/s=Δv
Answer:
The potential difference at the customer's house is 117.1 V.
Explanation:
a) The potential difference at the customer's house can be calculated as follows:
<u>Where</u>:
: is the potential difference at the customer's house
: is the potential difference from the main power lines = 120 V
: is the potential difference from the lines
The resistance, R, is:
Now, the potential difference at the customer's house is:
Therefore, the potential difference at the customer's house is 117.1 V.
I hope it helps you!
Answer:
18.2 square meters
Explanation:
If, on average, the solar panel receive 320W/m2 on a 24-hour day at 16.7% efficiency, then the actual daily energy it receive per square meter is:
So in order to match the daily energy provided by fossil, then the area of solar panel needed to install is