B
Force=Mass times acceleration
15=3x
x=5ms^2
When a swimmer pushes threw water to swim they are propelled forward because of the water resistance against the hand and feet.
Answer:
11.714 kW
Explanation:
Here is the complete question
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 34.0∘ above the horizontal. The car accelerates uniformly to a speed of 2.25 m/s in 10.5 s and then continues at constant speed. What power must the winch motor provide when the car is moving at constant speed?
Solution
Since the loaded ore car moves along the mine shaft at an angle of θ = 34° to the horizontal, if F is the force exerted on the cable, then the net force on the laoded ore car is F - mgsinθ = ma where mgsinθ = component of the car's weight along the incline, m = mass of loaded ore car = 950 kg and a = acceleration
F = m(a + gsinθ)
When the car is moving at constant speed, a = 0
So F = m(a + gsinθ) = F = 950(0 + 9.8sin34) = 5206.1 N
Since it continues at a constant speed of v = 2.25 m/s, the power of the winch motor is P = Fv = 5206.1 N × 2.25 m/s = 11713.7 W = 11.714 kW
Answer:
Explanation:
a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.
b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or 
c. The period, or T, of a wave is found in the equation
were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:
so
and
T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)
d. The speed of the wave is found in the equation
and since we already have the frequency and we solved for the wavelength already, filling in:
and
v = 50.0(20.0) so
v = 1.00 × 10³ m/s
And there you go!
We can rearrange the mirror equation before plugging our values in.
1/p = 1/f - 1/q.
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p <-- cross multiplication
13.33cm = p
Now that we have the value of p, we can plug it into the magnification equation.
M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'
So the height of the image produced by the mirror is 9.6cm.
