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VladimirAG [237]
2 years ago
15

Wind gusts create ripples on the ocean that have a wavelength of 2.23 cm and propagate at 4.12 m/s. What is their frequency (in

Hz)?
Physics
1 answer:
stellarik [79]2 years ago
8 0

Answer:

The frequency of the wave is 184.75 Hz.

Explanation:

Given;

wavelength of the wind, λ = 2.23 cm = 0.0223 m

speed of the wind wave, v = 4.12 m/s

the frequency of the wave = ?

The frequency of the wave is calculated as;

F = V / λ

Where;

F is the frequency of the wave

Substitute the given values and solve for F

F = (4.12) / (0.0223)

F = 184.75 Hz.

Therefore, the frequency of the wave is 184.75 Hz.

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What is the force on an object that goes from 35 m/s to 85 m/s in 20 seconds and has a mass of 148 kg
Sever21 [200]
F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N
3 0
3 years ago
The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I
erma4kov [3.2K]

Answer:

Part a)

I_{end} = \frac{mL^2}{3}

Part b)

I_{edge} = \frac{2ma^2}{3}

Explanation:

As we know that by parallel axis theorem we will have

I_p = I_{cm} + Md^2

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

I = \frac{mL^2}{12}

now if we need to find the inertia from its end then we will have

I_{end} = I_{cm} + Md^2

I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2

I_{end} = \frac{mL^2}{3}

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

I = \frac{ma^2}{6}

now if we need to find the inertia about an axis passing through its edge

I_{edge} = I_{cm} + Md^2

I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2

I_{edge} = \frac{2ma^2}{3}

7 0
3 years ago
If two swimmers compete in a race, does the faster swimmer develop more power?
valkas [14]
Power is equal to energy per unit time. In this case, power is proportional to energy while is inversely proportional to time,on the other hand. Given the two swimmers exerts same amount of energy but the faster swimmer just does things in faster time, then the faster swimmer should develop more power from shorter time
7 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
3 years ago
Assuming 84.0% efficiency for the conversion of electrical power by the motor, what current must the 13.0-V batteries of a 716 k
tino4ka555 [31]

Answer:

\mathbf{ current(I) =1766.67 \ A}

Explanation:

Given that:

The air resistance and friction = 700 N

The gravity caused force = 716 × 9.8 = 7016.8

Total force = (7016.8 + 700) N

Total force = 7716.8 N

∴

13 \times  current(I) \times 0.84 = \dfrac{7716.8 \times 300}{2 \times 60}

current(I) \times 10.92= 19292

current(I) = \dfrac{19292}{10.92}

\mathbf{ current(I) =1766.67 \ A}

8 0
2 years ago
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