Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p =
vμω²A²
p =
√(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz
Answer:
The distance is 6259.31 meters.
Explanation:
We shall use the Reyligh criterion to solve the problem
For diffraction due to circular aperture we have
Assuming that human eye is circular we have


Applying the given values we have

Answer:
ΔLa/ΔLb = 1
Explanation:
The change in length of a solid is given by the following formula:
ΔL = α L ΔT
where,
ΔL = Change in length
α = coefficient of linear expansion
L = Original Length
ΔT = Change in Temperature
Since, the length and change in temperature for both rods are same. Also, the material of each rod is same, which implies that coefficient of linear expansion for both rods is same. Hence, the ratio of change in length of both rods will be:
<u>ΔLa/ΔLb = 1</u>
Answer:
A. <u>The water decreases the friction between the floor and the feet </u>
Explanation: Think about it like this, when your in the shower and water is on your skin, you can scrub it fluidly, but when the water dries in your towel, there is more friction when your rub your skin, this is because the molecules in water aren't as compact as solids, so anything acting against it, is most likely to disperse from it.
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N