In the circular motion of the hammer, the centripetal force is given by
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
Answer:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
To help you I need to assume a wavelength and then calculate the momentum.
The momentum equation for photons is:
p = h / λ , this is the division of Plank's constant by the wavelength.
Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:
p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s
p = 1.01067 * 10^ - 27 kg*m/s which must be rounded to three significant figures.
With that, p = 1.01 * 10 ^ -27 kg*m/s
The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s
So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.
Answer: 0.98m
Explanation:
P = -74 mm Hg = 9605 Pa = 9709N/m^2
= 9605 kg m/s^2/m^2
density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3
Pressure equation: P = rho g h
h = P/(rho g)
h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)
h = 0.98 m
0.98m is the maximum depth he could have been.
Weight = (mass) x (gravity)
Acceleration of gravity on Earth = 9.8 m/s²
Weight on Earth = (mass) x (9.8 m/s²)
Divide each side by (9.8 m/s²): Mass = (weight) / (9.8 m/s²)
Mass = (650 N) / (9.8 m/s²)
Mass = 66.33 kg (rounded)
