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Mkey [24]
3 years ago
5

How far does light travel in the time it takes sound to travel 1 cm in air at 20°c?

Physics
1 answer:
kykrilka [37]3 years ago
8 0
The speed of sound at 20^{\circ}C is approximately v=343 m/s. The distance covered by the sound wave is
s=1 cm=0.01 m
And the time it takes is
t= \frac{S}{v}= \frac{0.01 m}{343 m/s}=2.9 \cdot 10^{-5} s

Now we want to find how far the light travels during this time. Light travels at speed c=3 \cdot 10^8 m/s, therefore the distance it covers during this time is
S=ct = (3 \cdot 10^8 m/s)(2.9 \cdot 10^{-5} s)=8700 m
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Determine the largest intensity w of the uniform loading that can be applied to the frame without causing either the average nor
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two ends of an inextensible string of length 12m are attached to points Aand B 1.2m apart,in the same horizontal plane.a mass 20
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3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
blondinia [14]

Answer:

1.2 m/s

0.31 m

0.15 m

Explanation:

Time period is

T=2.5\times 2\\\Rightarrow T=5\ s

Frequency is

f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{5}\\\Rightarrow f=0.2\ Hz

Velocity is given by

v=f\lambda\\\Rightarrow v=0.2\times 6\\\Rightarrow v=1.2\ m/s

The waves are traveling at 1.2 m/s

Amplitude is given by

A=\dfrac{d}{2}\\\Rightarrow A=\dfrac{0.62}{2}\\\Rightarrow A=0.31\ m

Amplitude is 0.31 m

If d = 0.3 m

A=\dfrac{0.3}{2}=0.15\ m

The amplitude would be 0.15 m. The speed would remain the same.

8 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

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3 years ago
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