What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di
1 answer:
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The formula for the period of wave is: wave period is equals to 1 over the frequency.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
Answer:
a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r
Explanation:
The law of universal gravitation is
F = G m M / r²
Part A
Let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / R₂
G m M / R₂² = m v² / R₂
v² = G M / R₂
They give us the density of the planet
ρ = M / V
V = 4/3 π R₁³
M = ρ V
M = ρ 4/3 π R₁³
v² = 4/3 π G ρ R₁³ / R₂
K = ½ m v²
K = ½ m (4/3 π G ρ R₁³ / R₂)
K = 2/3 π G m ρ R₁³ / R₂
Part B
Potential energy and strength are related
F = - dU / dr
∫ dU = - ∫ F. dr
The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1
U- U₀ = G m M ∫ dr / r²
U - U₀ = G m M (- r⁻¹)
We evaluate for
U - U₀ = -G m M (1 / - 1 /
)
They indicate that for ri = ∞ U₀ = 0
U = - G m M / r