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3 years ago

What do you understand if the frequency of a sound is 70 Hz?

Physics

1 answer:

ankoles [38]3 years ago

4 0

Answer:

see below

Explanation:

frequency of a wave is the number of complete waves produced per second

so 70 Hz

means, that 70 complete waves are produced in one second,

hope this helps

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A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 m/s in the positive direction of

Burka [1]

Answer:

0.8m/s

Explanation:

Weight of mas,F=763 N

Mass of man= $\frac{F}{g}=\frac{763}{9.8}=77.86 kg$

By using $g=9.8m/s^2$

Weight of flatcar=F'=3513 N

Mass of flatcar= $\frac{3513}{9.8}=358.5 Kg$

Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg

Velocity of system=19.8m/s

Let v be the velocity of flatcar with respect to ground

Velocity of man relative to the flatcar= $-4.68m/s$

Final velocity of man with respect to ground=v-4.68

By using law of conservation of momentum

Initial momentum=Momentum of car+momentum of flatcar

$436.36(19.8)=77.86(v-4.68)+358.5v$

$8639.928=77.86v-364.3848+358.5v$

$8639.928+364.3848=436.36 v$

$9004.3128=436.36v$

$v=\frac{9004.3128}{436.36}$

$v=20.6 m/s$

Initial speed of flatcar=Speed of system

Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s

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3 years ago

2. A person standing at

elena-s [515]

God is good man what can you say but its 18.66x30301=362728

7 0

3 years ago

А is a push or pull.

guajiro [1.7K]

Answer:

Explanation:

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3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

7. How do Newton’s laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can pass a

Arisa [49]

The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down

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3 years ago

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What do you understand if the frequency of a sound is 70 Hz? ​

What do you understand if the frequency of a sound is 70 Hz?