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Inga [223]
3 years ago
15

A hammer taps on the end of a 5.0-mm-long metal bar at room temperature. A microphone at the other end of the bar picks up two p

ulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 8.40 msms .What is the speed of sound in this metal?
Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Answer:

810.37 m/s

Explanation:

s = Displacement = 5 m

v_a = Speed of sound in air = 343 m/s

Time taken to travel the distance

t'=\dfrac{s}{v_a}\\\Rightarrow t'=\dfrac{5}{343}\\\Rightarrow t'=0.01457\ s

Time for the sound in metal

t=0.01457-8.4\times 10^{-3}=0.00617\ s

Speed of sound is given by

v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5}{0.00617}\\\Rightarrow v=810.37277\ m/s

The speed of sound in the metal is 810.37 m/s

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Answer:

The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.

Explanation:

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For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
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Answer rain gauge measures rain shadow units millimetres

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2 years ago
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The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the
ratelena [41]

Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0
3 years ago
0.0084x91 ...................
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Answer:

0.0084×91=0.7644

Explanation:

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