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4 years ago

A jet with mass m = 90000.0 kg jet accelerates down the runway for takeoff at 1.6 m/s2.

1 answer:

6 0

This question requires the use of the equation of motion:
v = u + at [v is final velocity (0), u is initial velocity (24), a is acceleration, t is time (13)]
to calculate the acceleration. This can then be multiplied by the mass of the plane to obtain the net force via:
F = ma (F is force, m is mass, a is acceleration)
First, we calculate the acceleration:
0 = 24 + 13(a)
a = -24/13 m/s^2
The force is then:
F = 90000 * (-24/13)
F = -1.66*10^5 Newtons
The negative sign indicates that the force and acceleration are in the opposite direction as the velocity (since we took velocity to be positive)

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of

IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

$T_2 = 1.008$ % higher than $T_1$

$T_2 = 0.99$ % lower than $T_1$

Explanation:

From the question we are told that

The first string has a frequency of $f_1 = 230 Hz$

The period of the beat is $t_{beat} = 0.99s$

Generally the frequency of the beat is

$f_{beat} = \frac{1}{t_{beat}}$

Substituting values

$f_{beat} = \frac{1}{0.99}$

$= 1.01 Hz$

From the question

$f_2 - f_1 = f_{beat}$ for $f_2$ having a higher tension

$f_2 - 230 = 1.01$

$f_2 = 231.01Hz$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}$

$T_2 = 1.008$ % higher than $T_1$

For $f_2$ having a lower tension

$f_1 - f_2 = f_{beat}$

$230 - f_2 = 1.01$

$f_2 = 230 -1.01$

$= 228.99$

From the question

$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }$

$\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}$

Substituting values

$\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}$

$T_2 = 0.99$ % lower than $T_1$

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