Answer:
(a) T = 0.412s
(b) f = 2.42Hz
(c) w = 15.25 rad/s
(d) k = 86.75N/m
(e) vmax = 5.03 m/s
Explanation:
Given information:
m: mass of the block = 0.373kg
A: amplitude of oscillation = 22cm = 0.22m
T: period of oscillation = 0.412s
(a) The period is the time of one complete oscillation = 0.412s
The period is 0.412s
(b) The frequency is calculated by using the following formula:

The frequency is 2.42 Hz
(c) The angular frequency is:

The angular frequency is 15.25 rad/s
(d) The spring constant is calculated by solving the following equation for k:

The spring constant is 86.75N/m
(e) The maximum speed is:

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

The maximum force that the spring exerts on the block is 17.35N
The solution is:tan(θ) = opp / adj tan(θ) = y/x xtan(θ) = y
Find x:
x = y/tan(θ)
So x = 3/tan(π/6)
Perform implicit differentiation to get the equation:
dx/dt * tan(θ) + x * sec²(θ) * dθ/dt = dy/dt
Since altitude remains the same, dy/dt = 0. Now...
dx/dt * tan(π/6) + 3/tan(π/6) * sec²(π/4) * -π/4 = 0
changing the equation, will give us:
dx/dt = [3/tan(π/6) * sec²(π/6) * π/4} / tan(π/6) ≈ 12.83 km/min
Answer:
P = 180 [w]
Explanation:
To solve this problem we must use ohm's law, which is defined by the following formula.
V = I*R & P = V*I
where:
V = voltage = 200[volts]
I = current [amp]
R = resistance [ohm]
P = power [watts]
Since the bulbs are connected in series, the powers should be summed
P = 60 + 60 + 60
P = 180 [watts]
Now we can calculate the current
I = 180/200
I = 0.9[amp]
Attached is an image where we see the three bulbs connected in series, in the circuit we see that the current is the same for all the elements connected to the circuit.
And the power is defined by P = V*I
we know that the voltage is equal to 200[V], therefore
P = 200*0.9
P = 180 [w]
Answer:
two wavelength are present
as wavelength is counted either from crest to crest or trough to trough and there are only 2
The girls distance is 50 km in 2 hours.