Answer: 750 N
Explanation:
The net force is 1200 - 450 = 750 N
As we are told the speed is constant, then this force must be increasing the car's potential energy by climbing a hill.
F = mgsinθ
If we knew the car mass, we could find the hill slope angle.
If we knew the hill slope angle, we could find the car mass.
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
R=4.22*10⁴km
Explanation:
The tangential speed 
of the geosynchronous satellite is given by:
Because 
is the circumference length (the distance traveled) and T is the period (the interval of time).
Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:
If we substitute the expression for in this formula, we get:
Since the centripetal force is the gravitational force 
between the satellite and the Earth, we know that:
![F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }](https://tex.z-dn.net/?f=F_g%3D%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%5C%5C%5C%5C%5Cimplies%20%5Cfrac%7BGMm%7D%7BR%5E%7B2%7D%7D%3D%5Cfrac%7B4m%5Cpi%20%5E%7B2%7DR%7D%7BT%5E%7B2%7D%7D%5C%5C%5C%5CR%5E%7B3%7D%3D%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%5C%5C%5C%5CR%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGMT%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%20%7D)
Where G is the gravitational constant (
) and M is the mass of the Earth (
). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:
![R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%2A10%5E%7B-11%7DNm%5E%7B2%7D%2Fkg%5E%7B2%7D%29%285.97%2A10%5E%7B24%7Dkg%29%2886400s%29%5E%7B2%7D%7D%7B4%5Cpi%5E%7B2%7D%7D%7D%5C%5C%5C%5CR%3D4.22%2A10%5E%7B7%7Dm%3D4.22%2A10%5E%7B4%7Dkm)
This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.
You haven't told us what the passing percentage is on the exam,
or what the passing percentage is for the semester, or any of that.
The chemical formular for water is H2O.
The H aspect of the formula stands for hydrogen gas and the subscript 2 which is attached to the H symbol signifies that two atoms of hydrogen are joined together, that is two atom of hydrogen are present.
The chemical formula of water indicates that, two atom of hydrogen react with one atom of oxygen to form one molecule of water.
In chemical formulae, subscripts are normally used to indicate the number of atoms that are present in a molecule.
