The answer is <span>A. Speed=100 million m/s and frequency = 50 million Hz.</span>
Let's calculate for each choice the wavelength using the equation:
v = f × λ ⇒ λ = v ÷ f<span>
where:
v - the speed,
f - the frequency,
</span>λ - the wavelength.
A:
v = 100 000 000 m/s
f = 50 000 000 Hz = 50 000 000 1/s (Since f = 1/T, so units are Hz = 1/s)
⇒ λ = 100 000 000 ÷ 50 000 000 = 2 m
B:
v = 150 000 000 m/s
f = 1 500 Hz = 1 500 1/s
⇒ λ = 150 000 000 m/s ÷ 1 500 = 100 000 m
B:
v = 300 000 000 m/s
f = 100 Hz = 100 1/s
⇒ λ = 300 000 000 m/s ÷ 100 = 3 000 000 m
According to these calculations, the shortest wavelength is needed for choice A.
Oscillatory motion.(also known as Periodic motion)
Hope it helps...
Regards,
Leukonov/Olegion.
Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
