<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>
13 cm
just solved the same question on mastering physics<span />
Answer:
Explanation:
Calculating the exit temperature for K = 1.4
The value of 
is determined via the expression:
where ;
R = universal gas constant = 
k = constant = 1.4
The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :
------ equation(1)
we can rewrite the above equation as :
where:
Thus, the exit temperature = 402.36 K
The exit pressure is determined by using the relation:
Therefore, the exit pressure is 17.79 bar
Answer:
A Fan Cart Initially Has An Acceleration Of 1.6m/s/s When It's Fan Is Directed Straight Backwards. If You Rotate The Fan By 45o, By What Percentage Do You Expect The Fan Cart's Thrust To Decrease? (Answer Should Be In Units Of 96)
a. 45%
b. 29%
c. 71%
d. 50%
The correct answer is d.
d. 50%
Explanation:
Fan cart acceleration = 1.6 m/s²
Thrust = 0.25×π×D²×ρ×v×Δv
where Δv = acceleration component and all factors remaining cconstant, when the fan is rotated by 45 ° the diameter changes to D₂ = sin 45 ×D
or 0.707×D. The thrust becomes 0.25×π×(0.707×D)²×ρ×v×Δv
=0.25×π×0.5×D²×ρ×v×Δv or 0.5(0.25×π×D²×ρ×v×Δv)
That is the thrust reduces by 50 %
