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babunello [35]
3 years ago
9

The frequency of a wave is 1) A) measured in cycles per second. B) measured in hertz (Hz). C) the number of peaks passing by any

point each second. D) equal to the speed of the wave divided by the wavelength of the wave. E) all of the above
Physics
1 answer:
7nadin3 [17]3 years ago
6 0

Answer: E) all of the above

Explanation:

Frequency is the number of  vibrations in one second. It is also defined as the number of crests that pass a point in a given time.

The frequency and the wavelength has an inverse relationship.

Frequency is measured in cycles per second or Hertz(Hz).

The relationship between wavelength and frequency of the wave follows the equation:

\nu=\frac{c}{\lambda}

where,

\nu = frequency of the wave

c = speed of light  

\lambda = wavelength of the wave

Thus all the given statements are true.

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A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
avanturin [10]

Answer:

(a) W= 44N

(b)W= 31.65 N

Explanation:

Data

T=44 N : Maximum force that the rope can withstand without breaking

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a)  We apply the formula (1) at constant speed , then, a=0

W: heaviest fish that can be pulled up vertically

∑F = 0

T-W =0  

W = T

W= 44N

(b)  We apply the formula (1) , a= 1.26 m/s²

W: heaviest fish that can be pulled up vertically

W= m*g

m= W/g

g= 9.8 m/s² : acceleration due to gravity

∑F = 0

T-W = m*a

T= W+(W/g)*a

44=W*(1+1/9.8)* (1.26 )

44= W* 1.39

W= 44/1.39

W= 31.65 N

7 0
3 years ago
An advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers (n = 3.50
sasho [114]

Answer:

d = 6.43 cm

Explanation:

Given:

- Speed resistance coefficient in silicon n = 3.50

- Memory takes processing time t_p = 0.50 ns

- Information is to be obtained within T = 2.0 ns

Find:

- What is the maximum distance the memory unit can be from the central processing unit?

Solution:

- The amount of time taken for information pulse to travel to memory unit:

                            t_m = T - t_p

                            t_m = 2.0 - 0.5 = 1.5 ns

- We will use a basic relationship for distance traveled with respect to speed of light and time:

                           d = V*t_m

- Where speed of light in silicon medium is given by:

                           V = c / n

- Hence,              d = c*t_m / n

-Evaluate:           d = 3*10^8*1.5*10^-9 / 3.50

                           d = 0.129 m 12.9 cm

- The above is the distance for pulse going to and fro the memory and central unit. So the distance between the two is actually d / 2 = 6.43 cm

8 0
3 years ago
Which two options describe physical properties of matter?
Artyom0805 [142]

Answer:

A and E.

Explanation:

Physical properties have to do with things that are not done chemically.

A has to do with light that you can see.

B has to do with Ph (If not A this is your next answer)

C mentions a Patina which is a chemical reaction known as oxidation

D has to do with chemical reactions

E is always correct.  One of the fundamental laws about matter is that it must always have mass.

I am 99% confident in A and E as your answer, but if it is wrong go with B and E.

Hope this helps!

7 0
3 years ago
If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co
sergij07 [2.7K]

Answer:

Angular velocity, \omega=35.36\ rad/s

Explanation:

It is given that,

Maximum emf generated in the coil, \epsilon=20\ V

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, B=9\times 10^{-3}\ T

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}

\omega=35.36\ rad/s

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

6 0
3 years ago
Please help me↓
olga2289 [7]

Answer:

unmmm u can say that it is going mt 15s

3 0
3 years ago
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