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3 years ago

Please help me it's URGENT (#10 btw and also how do I solve for time

Physics

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

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Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

3 years ago

The more ______ an object moves, the more __________ it has (Kinetic Energy)

belka [17]

Answer:

__________

Explanation:

3 0

3 years ago

Question 9

Ilya [14]

True ! Attending the funeral of a foreign leader IS fulfilling the role of head of state.

5 0

3 years ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

umka2103 [35]

Answer:

a) $F=2.048\times 10^{-7}\ N$

b) $a=0.1138\ m.s^{-2}$

Explanation:

Given:

mass of raindrops, $m=1.8\times 10^{-6}\ kg$

charge on the raindrops, $q=+21\times 10^{-12}\ C$

horizontal distance between the raindrops, $r=0.0044\ m$

From the Coulomb's Law the force between the charges is given as:

$F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}$

we have:

$\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}$

Now force:

$F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}$

$F=2.048\times 10^{-7}\ N$

Now the acceleration on the raindrops due to the electrostatic force:

$a=\frac{F}{m}$

$a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}$

$a=0.1138\ m.s^{-2}$

7 0

3 years ago

Read 2 more answers

A car weighing 14,700 N is speeding down a highway with a velocity of 99 km/h. What is the

tankabanditka [31]

Answer: 148348.6239 kg•m/s

Explanation: Firstly, we need to convert the 14700 N into kilograms, and to do so, use the formula net force is equal to mass times acceleration and rearrange the formula to find mass like shown below...

F = ma

F/a = m

14700/9.81 = 1498.470948 kg, this is your mass

Now that we convert it into kilograms, plug all the numbers into the variable of the momentum formula.

Momentum formula is P = mass x velocity

Like this:

P = 1498.470948 x 99

p = 148348.6239 kg•m/s.

I believe that is your answer, hope that helps you even a bit out.

Thanks.

7 0

2 years ago

Please help me it's URGENT (#10 btw and also how do I solve for time​

Please help me it's URGENT (#10 btw and also how do I solve for time