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3 years ago

When air is blown across the top of an open

1 answer:

8 0

When air is blown across the top of an open water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced in a process called "refraction".

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A wrestler weighs in for the first match on the moon. will the athlete weigh more or less on the moon than he does on Earth?

Firdavs [7]

The moon is 1/4 the size of Earth, so the moon's gravity is much less than the earth's gravity, 83.3% (or 5/6) less to be exact. Finally, "weight" is a measure of the gravitational pull between two objects. So of course you would weigh much less on the moon.

4 0

3 years ago

How much kinetic energy does a 50 kg object have if its moving at a velocity of 2 m/s?

bulgar [2K]

Answer:

C) 100 joules

Explanation:

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where m is the mass of the object and v its speed.

In this problem, we have an object of mass m = 50 kg and v = 2 m/s, so by using the formula we can find its kinetic energy:

$K=\frac{1}{2}(50 kg)(2 m/s)^2=100 J$

3 0

2 years ago

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0

3 years ago

Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER

lukranit [14]

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

3 0

2 years ago

What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?

givi [52]

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

6 0

2 years ago