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3 years ago

11

What best describes a sound wave? transverse and electromagnetic longitudinal and mechanical longitudinal and electromagnetic tr

ansverse and mechanical

2 answers:

GalinKa [24]3 years ago

7 0

Sound needs medium to travel and it can not travel without medium

so sound wave is a travelling wave

now we also know that sound wave propagate in form of rarefaction and compression.

So all medium particles travel in the direction of wave only

so it is a longitudinal wave also

so correct answer will be

<em>mechanical longitudinal </em>

ValentinkaMS [17]3 years ago

5 0

Answer:

longitudinal and mechanical

Explanation:

It's right on Edge

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A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp

AlekseyPX

Answer:

(A) 2.4 N-m

(B) $0.035kgm^2$

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque $\tau =Fr$ , here F is force and r is distance

So $\tau =20\times 0.12=2.4Nm$

(B) Moment of inertia is equal to $I=\frac{1}{2}mr^2$

So $I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2$

Torque is equal to $\tau =I\alpha$

So angular acceleration $\alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2$

(C) As the disk starts from rest

So initial angular speed $\omega _{0}=0rad/sec$

Time t = 4.6 sec

From first equation of motion we know that $\omega =\omega _0+\alpha t$

So $\omega =0+68.571\times 4.6=315.426rad/sec$

(D) Kinetic energy is equal to $KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J$

(E) From second equation of motion

$\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad$

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3 years ago

Lenny loves physics and math. In which Energy career pathway would these interests be the most helpful?

vazorg [7]

Answer:

it is D

Explanation:

cause i took the test

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3 years ago

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calculate the temperature at which the reading on Fahrenheit scale is equal to half the reading of Celsius scale

RideAnS [48]

Answer:

Therefore, the temperature at which the Fahrenheit scale reading is equal to half of the Celsius scale is −24.6∘C .

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2 years ago

A box (m = 20 kg) is sliding on a horizontal surface. it is connected to a massless hook by a light string passing over a massle

Varvara68 [4.7K]

mass of the box = 20 kg

force of friction on the box due to surface

$F_s = \mu_s N$

$F_s = 0.80 * 20 * 9.8$

$F_s = 156.8 N$

similarly kinetic friction on it

$F_k = \mu_k N$

$F_k = 0.30* 20 * 9.8$

$F_k = 58.8 N$

now the weight of the suspended block will be

$W = mg = 15*9.8 = 147 N$

so here the weight of the suspended block is less than the limiting friction on it

So here we will say that friction will counter balance the weight of the suspended block and it will not move at all

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3 years ago

A baseball is thrown straight up. the drag force is proportional to v2. part a in terms of g, what is the y-component of the bal

steposvetlana [31]

At the position of terminal speed the net acceleration of the ball will become zero

As we know that terminal speed will always reach when net force on the ball is zero and its speed will become constant.

So here at this position we can say

$F_d = F_g$

$kv^2 = mg$

$v =\sqrt{\frac{mg}{k}}$

now when ball is moving at half of the terminal speed in upward direction then net force on the ball in downwards direction will be

$F_{net} = mg + F_d$

$F_{net} = mg + kv^2$

here speed of the ball is half of the terminal speed

$v = \frac{1}{2}*\sqrt{\frac{mg}{k}}$

then we have

$F_{net} = mg + k*\frac{mg}{4k}$

$F_{net} = \frac{5mg}{4}$

now acceleration will be given as

$a = \frac{F_{net}}{m}$

now we have

$a = \frac{5g}{4}$

$a = 12.25 m/s^2$ downwards

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3 years ago