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Zolol [24]
3 years ago
11

If the amplitude of a wave is doubled, it has half the velocity. TRUE or FALSE?

Physics
1 answer:
blondinia [14]3 years ago
7 0

Answer:

true

Explanation:

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How is water vapour different from a mixture of hydrogen and oxygen atoms?
Pavel [41]

Answer: in water vapor hydrogen atoms are bonded to oxygen atoms.  In a mixture of hydrogen and oxygen hydrogen atoms are bonded to hydrogen atoms and oxygen atoms are bonded to oxygen atoms.

Explanation:

5 0
3 years ago
What is the wavelength of an electromagnetic wave that travels at 3 10 m/s and has a frequency of 60 mhz? (1 mhz = 1,000,000 hz)
Dahasolnce [82]
V = 310 m/s
f = 60 MHz = 60 × 10^6 Hz
v = xf
x = v/f
x = 310/(60 × 10^6) m
x = 5.166667 × 10^(−6) m
8 0
3 years ago
Read 2 more answers
What is the current
Art [367]

Answer:

2.5mA

Explanation:

In the picture.

I think it's a clear.

8 0
3 years ago
What will be the ME of a machine that produces a 240 N work with a 300
Elden [556K]

Answer:

Efficiency = 80%

Explanation:

Given the following data;

Work output = 240 N

Work Input = 300 N

To find the mechanical efficiency of a machine;

Efficiency = \frac {Out-put \; work}{In-put \; work} * 100

Substituting into the equation, we have;

Efficiency = \frac {240}{300} * 100

Efficiency = 0.8 * 100

Efficiency = 80%

Therefore, the mechanical efficiency of the machine is 80 percent.

3 0
3 years ago
Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

6 0
3 years ago
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