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2 years ago

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

Physics

1 answer:

NNADVOKAT [17]2 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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A measuring microscope is used to examine the interference pattern. It is found that the average distance between the centers of

diamong [38]

Answer:

2n t = m λ₀ , R = 0.240 mm

Explanation:

The interference by regency in thin films uses two rays mainly the one reflected on the surface and the one reflected on the inside of the film.

The ray that is reflected in the upper part of the film has a phase change of 180º since the ray stops from a medium with a low refractive index to one with a higher regrading index,

-This phase change is the introduction of a λ/2 change

-The ray passing through the film has a change in wavelength due to the refractive index of the medium

λ₀ = λ / n

Therefore Taking into account this fact the destructive interference expression introduces an integer phase change, then the extra distance 2t is

2 t = (m’+ ½ + ½) λ₀ / n

2t = (m’+1) λ₀ / n

m = m’+ 1

2n t = m λ₀

With m = 0, 1, 2, ...

Where t is the thickness of the film, n the refractive index of the medium, λ the wavelength

The thickness of a hair is the thickness of the film t

2R = t

R = t / 2

R = 0480/2

R = 0.240 mm

3 0

1 year ago

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

Elan Coil [88]

Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.

Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement.

Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder.

Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

8 0

2 years ago

A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t

klasskru [66]

Answer:

a. $a. \ \frac{7}{25}$

$b.\ \ \ P(D_1D_2)=\frac{6}{25}$

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let $A_i=i^{th}$ photo is accepted and the probability $D_i=i^{th}$ is defected.

Therefore:

$P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}$

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

$P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}$

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0

2 years ago

If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac

maks197457 [2]

In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω₂:
ω₂ = (r₁/ r₂)² · ω₁

Since we know that r₁ = 4r₂, we get:
ω₂ = (4)² · ω₁
= 16 · ω₁

Hence, the protostar will be rotating 16 times faster.

5 0

2 years ago

A sleeping 68 kg man has a metabolic power of 79 w .

Lesechka [4]

65W * 8h * 3600s/h = 1.9e6 J = 447 Cal

3 0

2 years ago