For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
Answer:
<h3>62.5N</h3>
Explanation:
The pressure at one end of the piston is equal to the pressure on the second piston.
Pressure = Force/Area
F1/A1 = F2/A2
Given
F1 = 250N
A1 = 2.0m²
A2 = 0.5m²
F2 = ?
Substituting the given values in the formula;
250/2 = F2/0.5
cross multiply
250*0.5 = 2F2
125 = 2F2
F2 = 125/2
F2 = 62.5N
Hence the force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N
Answer:Orbital period =21.22hrs
Explanation:
given that
mass of earth M = 5.97 x 10^24 kg
radius of a satellite's orbit, R= earth's radius + height of the satellite
6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m
Speed of satellite, v= 
where G = 6.673 x 10-11 N m2/kg2
V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)
V =10,241082.2
v= 3,200.2m/s
a) Orbital period
= 
V= 
T= 2
r/ V
= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s
=76,385.1 s
60 sec= 1min
60mins = 1hr
76,385.1s =hr
76,385.1/3600=21.22hrs