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GarryVolchara [31]
4 years ago
10

If the period of a clock signal is 500 ps, the frequency is:_____

Physics
1 answer:
NNADVOKAT [17]4 years ago
4 0

Answer:

The frequency of the signal is 2 GHz

Explanation:

Given;

period of the clock signal, T = 500 ps = 500 x 10⁻¹² s

the frequency of the signal is given by;

F= \frac{1}{T}\\\\F = \frac{1}{500*10^{-12}}\\\\F = 2*10^{9} \ Hz

F = 2 GHz

Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz

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3 years ago
A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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3 years ago
An experiment is conducted to determine how the arm length of a pendulum effects its period. The experimental design calls for v
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8 0
3 years ago
Read 2 more answers
2. Two cars A and B are heading forward. The velocity of A and B are 30 m/s and 20 m/s
KengaRu [80]

Answer:

A.) 27000 kgm/s

18000 kgm/s

B.) Va = 22 m/s

C.) 19800 kgm/s

25200 kgm/s

Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g

M = 9×10^5/1000 = 900 kg

A.) Initial momentum of A

Mu = 900 × 30 = 27000 kgm/s

Initial momentum of B

Mu = 900 × 20 = 18000 kgm/s

B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.

Momentum before impact = momentum after impact

Given that Vb = 28 m/s

27000 + 18000 = 900Va + 900 × 28

45000 = 900Va + 25200

900Va = 45000 - 25200

900Va = 19800

Va = 19800/900

Va = 22 m/s

C.) Momentum of A after impact

MV = 900 × 22 = 19800 kgm/s

Momentum of B after impact

MV = 900 × 28 = 25200 kgm/s

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lianna [129]

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