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3 years ago

If a wave is traveling at 600 m/s and it's wavelength is 50 m, what is the frequency of the wave?

Physics

1 answer:

nydimaria [60]3 years ago

3 0

Answer: 12 Hertz

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Work Shown:

Frequency = (velocity)/(wave length)

Frequency = (600 m/s)/(50 m)

Frequency = (600/50) s^(-1)

Frequency = 12 Hertz

Note: 1 Hertz (abbreviated Hz) is one cycle per second. One way we can define what the cycle is, is to measure from one peak of the wave to the next neighboring peak. This is what the wavelength is. So saying "12 Hz" is basically saying "12 wavelengths per second".

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A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp

Lorico [155]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2$

b) Assume the grind stone is a solid disk, its moment of inertia is

$I = mR^2/2$

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

$I = 28*0.15^2/2 = 0.315 kgm^2$

So the friction torque is

$T_f = I\alpha = 0.315*0.21 = 0.06615 Nm$

The friction force is

$F_f = T_f/R = 0.06615 / 0.15 = 0.441 N$

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

$N = F_f/\mu = 0.441/0.2 = 2.205 N$

7 0

2 years ago

The force of gravity is the strongest of the four known force fields. True False

Setler79 [48]

The answer is true as gravity is powerful than any other force

8 0

2 years ago

Read 2 more answers

Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele

Marta_Voda [28]

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

8 0

2 years ago

When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin

saveliy_v [14]

True
If it helps you plz brainlest me

7 0

2 years ago