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atroni [7]
3 years ago
9

Radio waves from an fm station have a frequency of 103.1 mhz. if the waves travel with a speed of 3.00 ´ 10 m/s, what is the wav

elength?
Physics
1 answer:
ryzh [129]3 years ago
3 0
The frequency of the radio waves from the fm station is:
f=103.1 MHz = 103.1 \cdot 10^6 Hz
And the speed of the waves corresponds to the speed of light:
v=3 \cdot 10^8 m/s
Therefore, the wavelength of the radio waves can be found by using the following equation:
\lambda= \frac{v}{f}= \frac{3 \cdot 10^8 m/s}{103.1 \cdot 10^6 Hz}=2.91 m
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Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
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Answer:

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3 years ago
When large amplitude sound vibrates the ear significantly, a _______ results.
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D loud sound

Hope this helped:)
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3 years ago
Read 2 more answers
Learning Goal:
Rudiy27

Answer:

The questions are not complete so this is the complete questions

1. How much work W does the motor do on the platform during this process?

2. What is the angular velocity ωf of the platform at the end of this process?

3. What is the rotational kinetic energy, Ek, of the platform at the end of the process described above?

4. How long does it take for the motor to do the work done on the platform calculated in Part 1?

5. What is the average power delivered by the motor in the situation above?

6. . Note that the instantaneous power P delivered by the motor is directly proportional to ω, so P increases as the platform spins faster and faster. How does the instantaneous power P•f being delivered by the motor at the time t•f compare to the average power

P(average) calculated in Part e?

Explanation:

Given that,

The torque τ=25Nm

Moment of inertia I =50kgm²

The platform is initially at rest,

ω•i=0 rad/sec

Revolution the torque produce is 12

Then, θ=12 revolution

1 revolution=2πrad

So, θ=24πrad

1. Work done in a rotational motion is give as

W=τ•Δθ

Given that the τ=25Nm and the initial angular displacement is 0rad

The final angular displacement is 24πrad

Δθ =(θ2-θ1)

Δθ=24π-0

Δθ=24πrad

Then,

W=τ•Δθ

W=25(24π)

W=25×24π

W=1884.96J

To 4s.f, W=1885J

2. Final angular velocity ωf

Using the angular equation

ω•f²=ω•i²+2•α•Δθ

We need to get angular acceleration

The torque is given as

τ=I•α

Given that,

I is moment if inertia =50kgm²

τ=25Nm

α=τ/I

α= 25/50

α=0.5rad/s²

Now, using the angular acceleration

ω•f²=ω•i²+2•α•Δθ

ω•f²=0²+2×0.5×24π

ω•f²=0+75.398

ω•f²=75.398

ω•f=√75.398

ω•f=8.68 rad/sec.

3. We need to find rotational Kinetic energy and it is given as

K.E, = ½I•ω²

Given that, I=50kgm² and ω•f=8.68rad/sec

Then,

K.E, =½I•ω²

K.E, =½×50×8.68²

K.E, =1884.96J

To 4s.f,

K.E, =1885J

Which is the same as the work done by the motor.

4. Time taken to complete part 1,

Using the rotational equation

ω•f=ω•i+α•t

Since, ω•f=8.68 rad/sec and ω•i=0

And α=0.5rad/s²

Then,

ω•f=ω•i+α•t

8.68=0+0.5t

8.68=0.5t

Then, t=8.68/0.5

t=17.36secs

5. The average power of rotational motion is given as

P(average) =Workdone/timetaken

Since,

Work done =1884.96J

Time taken =17.36sec

P(average) =Workdone/timetaken

P(average)=1884.96/17.36

P(average)= 108.58Watts

To 4s.f

P(average)=108.6Watts

6. We need to find •, it is given as

• =τ•ωf

Given that, ω•f=8.68rad/sec, τ=25Nm

•=25×8.68

•=217Watts

Then, the ratio of • to P(average) is

Ratio = •/ P(average)

Ratio= 217/108.58

Ratio=1.9985

Then, the ratio is approximately 2

Ratio=2

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4 years ago
A hummingbird flies 1.2 m along a straight path at a height of 3.4 m above the ground. Upon spotting a flower below, the humming
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Answer:1.84

Explanation:

Given

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Now bird drops 1.4 m to hover in front of the flower

So bird net displacement is

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displacement=\sqrt{3.4}

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3 years ago
A fugitive tries to hop on a freight train traveling at a constant speed of 5.5 m/s. Just as an empty box car passes him, the fu
stiks02 [169]
Th equations to be used here are the following:

a = (v - v₀)/t
x = v₀t + 0.5at²

The speed of the fugitive is the sum of his own speed plus the speed of the train. Thus, 
v₀ = 0 + 5.5 m/s = 5.5 m/s
v = 8 m/s + 5.5 m/s = 13.5 m/s

a.) We use the first equation to determine time
2.5 m/s² = (13.5 m/s - 5.5 m/s)t
Solving for t,
t = 3.2 seconds

b.) We use the answer in a) and the 2nd equation:

x = (5.5 m/s)(3.2 s) + 0.5(2.5 m/s²)(3.2 s)²
x = 30.4 meters
3 0
3 years ago
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