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Paraphin [41]
3 years ago
6

An object with an initial velocity of 27 m/s has a final velocity of 39 m/s after 13 seconds.

Physics
1 answer:
grigory [225]3 years ago
8 0

Answer:

<h3>The answer is 0.92 m/s²</h3>

Explanation:

To find the acceleration of an object given it's initial and final velocity and time taken we use the formula

a =  \frac{v - u}{t}  \\

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

From the question

v = 39 m/s

u = 27 m/s

t = 13 s

We have

a =  \frac{39 - 27}{13}  =  \frac{12}{13}  \\  = 0.923076... \:  \:  \:

We have the final answer as

<h3>0.92 m/s²</h3>

Hope this helps you

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3 years ago
A 25-g string is stretched with a tension of 43 N between two fixed points 12 m apart. What is the frequency of the second harmo
Svetllana [295]

Answer:

The frequency of the second harmonic (2f_o) is 11.97 Hz.

Explanation:

Given;

mass of the string, m = 25 g = 0.025kg

tension on the string, T = 43 N

length of the string, L = 12 m

The speed of wave on the string is given as;

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

where;

μ is mass per unit length = 0.025 / 12 = 0.002083 kg/m

v = \sqrt{\frac{43}{0.002083} }\\\\v  = 143.678 \ m/s

The wavelength of the first harmonic wave is given as;

L = \frac{1}{2} \lambda _o\\\\\lambda _o = 2L \\\\\lambda _o = 2 \ \times \ 12\\\\\lambda _o = 24 \ m

The frequency of the first harmonic is given as;

f_o = \frac{v}{\lambda _o} = \frac{v}{2L} = \frac{143.678}{24} = 5.99 \ Hz\\\\

The wavelength of the second harmonic wave is given as;

L = \lambda_1 \\\\\lambda_1 = 12 \ m

The frequency of the second harmonic is given as;

f_1 = \frac{v}{\lambda _1} = \frac{143.678}{12} = 11.97 \ Hz = 2(\frac{v}{\lambda _0}) = 2f_o

Therefore, the frequency of the second harmonic (2f_o) is 11.97 Hz.

8 0
3 years ago
How long does it take for an 8 kg pumpkin to hit the ground if dropped from a height of 55 m.
Sati [7]

Answer:

t = 3.35 s

Explanation:

It is given that,

Mass of a pumpkin, m = 8 kg

It is dropped from a height of 55 m

We need to find the time taken by it to hit the ground.

Initial velocity of the pumpkin, u = 0

Using second equation of motion to find it as follows :

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2(55)}{9.8}} \\\\t=3.35\ s

So, it will take 3.35 seconds to hit the ground.

7 0
3 years ago
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