Answer:
voltage and zero
Explanation:
I don't know i just did the Edg question
Answer:
Explanation:
ω = √k/m = √(33.9/0.28) = 11 rad/s
(a) maximum speed of the oscillating mass
vmax = ωA = 11(0.05) = 0.55 m/s
(b) speed of the oscillating mass when the spring is compressed 1.5 cm
The portion of total energy that is not spring potential is kinetic
½kA² - ½kx² = ½mv²
v = √(k(A² - x²)/m) = √(33.9(0.05² - 0.015²)/0.28 = 0.52482... ≈ 0.52 m/s
(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position
Different wording, but same question as part (b) 0.52 m/s
(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m
The portion of total energy that is not kinetic is spring potential
½kA² - ½mv² = ½kx²
x = √(kA² - m(vmax/2)²) / k) = √(33.9(0.05²) - 0.28(0.55/2)²) / 33.9)
x = 0.043305...≈ 4.3 cm
Answer:
The estimated time of arrival at the destination Airport will be 1730 PST.
Explanation:
Given:
Time at which flight departs from airport = 1615 MST
Duration of flight the reach destination airport = 2 hour 15 minute.
We need to find the estimated time of arrival at the destination airport.
Solution:
Now Given:
The Destination Airport is located in Pacific standard time zone.
Now first we will find the Arrival of flight at destination airport in MST.
Now we can say that;
Arrival of flight at destination airport will be equal to sum of Time at which flight departs from airport and Duration of flight the reach destination airport.
Arrival of flight at destination airport in MST =
Now We know that;
Mountain standard time is 1 hour ahead then Pacific Standard time.
Now Arrival of flight at destination airport in MST = 1830 MST
Arrival of flight at destination airport in PST = 1730 PST.
Hence The estimated time of arrival at the destination Airport will be 1730 PST.
1. Frequency:
The frequency of a light wave is given by:
where
is the speed of light
is the wavelength of the wave
In this problem, we have light with wavelength
Substituting into the equation, we find the frequency:
2. Period:
The period of a wave is equal to the reciprocal of the frequency:
The frequency of this light wave is (found in the previous exercise), so the period is:
Vf = Vi + at
Vf = 0 + (9.80)(4)
Vf = 39.2 m/s down
Hope a monkey breaks Arnold’s leg!