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3 years ago

How do vibration produce sound?

2 answers:

5 0

Vibrations form sound by releasing compressional sound waves. For example, in a guitar, when chords are strung they vibrate and produce sound.

Paladinen [302]3 years ago

4 0

All vibrating bodies produce sounds in molecules of the medium they are as air,wood,water

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State one of Kepler's laws of gravitation

jekas [21]

All planets move around the sun

3 0

3 years ago

Read 2 more answers

Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou

belka [17]

The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year.
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather

4 0

2 years ago

Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose

marusya05 [52]

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= - .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

6 0

3 years ago

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

2 years ago

What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660 N

==>T× 0.7660 = 29.4

==> T= 38.38 N

8 0

2 years ago