The farther away an object is from the Sun the slower it orbits around it. The closer an object is from the Sun the faster it orbits around it.
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
Answer:
32 seconds
Explanation:
m1 = 80 kg
m2 = 10 kg
v2 = 5m/s
According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:

Since the space station is 20 meters away, the time taken to reach it is given by:

It takes you 32 seconds to reach the station.
A trough is the lowest point in a wave.
Answer:
The speed of the sound wave on the string is 545.78 m/s.
Explanation:
Given;
mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m
tension of the string, T = 1400 N
The speed of the sound wave on the string is given by;

where;
v is the speed of the sound wave on the string
Substitute the given values and solve for speed,v,

Therefore, the speed of the sound wave on the string is 545.78 m/s.