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likoan [24]
3 years ago
12

A string is vibrating between two posts as shown above. Students are to determine the speed of the wave within this string. They

have already measured the amount of time necessary for the wave to oscillate up and down. The students must also take what other measurements to determine the speed of the wave?
A. The distance between the two posts.
B. The amplitude of the wave
C. The tension in the string
D. The amplitude of the wave and the tension in the string
E. The distance between the two posts, the amplitude of the wave, and the tension in the string
Physics
1 answer:
kow [346]3 years ago
6 0

Answer:

The distance between two posts

Explanation:

As the students have already measured the amount of time necessary for the wave to oscillate up and down that is on the string between two posts,They have to measure the speed of the wave.

        ie  v=fλ

<em>They have to measure distance between the posts. </em>

Because wavelength is equal to the distance between the posts ie full length of string.

The frequency of the wave is calculated by

               f= 1/T

Where is the time period which the students have calculated ie the amount of time taken to oscillate up and down,

Thus the wave speed is calculated using formula

            v=λ/T

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If a rocket travels 35,200 miles in 2 hours, what is its speed?
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Answer:

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PLEASE HELP!! ITS URGENT!!!​
Natasha2012 [34]

Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

ΣPbefore = ΣPafter

(m_{1}*v_{1}) - F*t = (m_{1}*v_{2})

where:

m₁ = mass of the hammer = 0.15 [m/s]

v₁ = velocity of the hammer = 8 [m/s]

F = force [N] (units of Newtons)

t = time = 0.0015 [s]

v₂ = velocity of the hammer after the impact = 0

(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]

Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.

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PLEASE HELP!!!!! which statements correctly conpare the masses of protons,neutrons,and electrons​
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2 years ago
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
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