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3 years ago

What change takes place in a wave when the frequency of the wave increases?

1 answer:

4 0

The answer is C. As the frequency of the waves increases, a greater number of wavelengths pass a given point per second. From the wave formula, we see that there is an indirect relationship between frequency and the wavelength. thus, as the frequency increases the wavelength decreases resulting to a smaller distance between the waves which will show greater number of wavelengths between waves.

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Wolf spiders carry their young on their backs. How does this help in their survival?

Goshia [24]

It intimidates other spider because it makes them look bigger than they really are.

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3 years ago

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What happens when an immovable object meets an unstoppable force?”

Alisiya [41]

Answer:

Sorry to tell you this but there’s no answer for that

Explanation:

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3 years ago

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Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 mfrom the ba

hjlf

Answer:

Δt = 0.315s

Explanation:

To calculate the time difference, in which both fans hear the batterstrike, you first calculate the time which takes the sound to travel the distances to both fans:

$t_1=\frac{d_1}{v_s}$

$t_2=\frac{d_2}{v_s}$

d1: distance to the first fan = 18.3 m

d2: distance to the second fan = 127 m

vs: speed of sound = 345 m/s

You replace the values of the parameters to calculate t1 and t2:

$t_1=\frac{18.3m}{345m/s}=0.053s\\\\t_2=\frac{127m}{345m/s}=0.368s$

The difference in time will be:

$\Delta t =t_2-t_2=0.368s-0.053s=0.315s$

Hence, the time difference between hearing the sound at the location s of both fans is 0.315s

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3 years ago

Cause of a severe storm

n200080 [17]

Causes of a severe storm are:

High winds

wildfires

hail

A severe thunderstorm includes winds of 58 MPH or greater

6 0

4 years ago

At time t=0 a 2150 kg rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. This forc

amm1812

Explanation:

Given that,

Mass of the rocket, m = 2150 kg

At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :

$F=At^2$

Here F = 888.93 N when t = 1.25 s

$F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2$

The value of A is $568.91\ N/s^2$ .

(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :

$J=\int\limits {F{\cdot} dt}$

Limits will be from 2 s to 2+ 4 = 6 s

It implies :

$J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns$

(b) Impulse is also equal to the change in momentum as :

$J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s$

Hence, this is the required solution.

5 0

3 years ago