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andreyandreev [35.5K]
3 years ago
15

For a given wave IF frequency doubles the wavelength

Physics
1 answer:
motikmotik3 years ago
7 0

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

v=f\lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}

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4 0
2 years ago
A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is
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Answer:

60 kg m/s

Explanation:

Let a\;\; m/s^2 be the acceleration of the object.

As the acceleration of the object is constant, so

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Given that applied force, F=6.00 N,

From Newton's second law, we have

F= m\times a,

\Rightarrow F=\frac {m(v-u)}{t} [from equation (i)]

\Rightarrow Ft=m(v-u)

\Rightarrow Ft=mv-mu

\Rightarrow mv-mu=6\times 10 [given that time, t=10 s and F=6 N]

\Rightarrow mv-mu=60 kg \;m/s

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

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3 years ago
A student wanted to check her idea that a planet that is closer to the Sun receives more of the Sun’s energy than a planet that
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C. a clamp light with a 150-W lightbulb placed 25 cm from the radiometer

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because the temperature has to be set, meaning it can't change, and it should be a different distance away than the first one.

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When making a budget, you should prioritize ___________ before "__________." A. Needs/wants B. Wants/needs C. Electricity/Water
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Answer A

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3 years ago
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
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Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

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u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

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J_y=m(v\sin (55)-(-u))

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J_x=2\times 4.42\cos (55)=20.646

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J_{net}=\sqrt{(38.32)^2+(20.64)^2}

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at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
3 years ago
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