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CaHeK987 [17]
2 years ago
5

What is the frequency of a wave that has a period of vibration of 2 seconds?

Physics
1 answer:
34kurt2 years ago
7 0

Answer:

The answer is 0.5 Hz

Explanation:

Its pretty easy to get the answer. One hertz (Hz) is equal to one cycle or period per second. So, just divide the period by the number of seconds.

1 period/2 secs = 1/2 Hz or 0.5 Hz

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murzikaleks [220]

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4 0
3 years ago
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A(n)............ exerts a force on a system. It is symbolised as a subscript to the force.
Marta_Voda [28]
4- Push— is the answer
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3 years ago
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A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl
raketka [301]
The wavelength of the note is \lambda = 39.1 cm = 0.391 m. Since the speed of the wave is the speed of sound, c=344 m/s, the frequency of the note is
f= \frac{c}{\lambda}=879.8 Hz

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }
where \mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
L= \frac{1}{2f}  \sqrt{ \frac{T}{\mu} } =0.31 m
8 0
3 years ago
A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately
Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car V_{c} = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0
2 years ago
A coin is dropped in a 15.0 m deep well.
labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0
2 years ago
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