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3 years ago

Particles in a medium through which a sound wave is traveling

Physics

1 answer:

GuDViN [60]3 years ago

7 0

Answer:

EL SONIDO PUEDE VIAJAR POR CUALQUIER MEDIO MATERIAL, COMO

AIRE, NITRÓGENO, AGUA, TITANIO, ETC... ASÍ QUE EL

SONIDO VIAJA A TRAVÉS DE MOLÉCULAS.

Explanation:

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A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un

Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0

3 years ago

Using parallax, what is the distance to the<br> farthest stars you can measure?

aev [14]

Parallax angles of less than 0.01 arcsec are very difficult to measure from Earth because of the effects of the Earth's atmosphere. This limits Earth based telescopes to measuring the distances to stars about 1/0.01 or 100 parsecs away.

6 0

3 years ago

A ball of mass 0.15 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. b)

Art [367]

Answer:

Explanation:

If v be the velocity just after the rebound

Kinetic energy will be converted into potential energy

1/2 m v² = mgh

v² = 2gh

v = √ 2gh

= √ 2 x 9.8 x .96

= 4.33 m / s

4 0

3 years ago

A man stands at the edge of a cliff and throws a rock downward with A speed of 12.0 m s . Sometimes later it strikes the ground

kvasek [131]

B) 48.0 m/s

We can actually start to solve the problem from B for simplicity.

The motion of the rock is a uniformly accelerated motion (free fall), so we can find the final speed using the following suvat equation

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

where

$v$ is the final velocity

$u = 12.0$ is the initial velocity (positive since we take downward as positive direction)

$a=g=9.8 m/s^2$ is the acceleration of gravity

s = 110 m is the vertical displacement

Solving for v, we find the final velocity (and so, the speed of the rock at impact):

$v = \sqrt{u^2+2as}=\sqrt{12^2+2(9.8)(110)}=48.0 m/s$

A) 3.67 s

Now we can find the time of flight of the rock by using the following suvat equation

$v=u+at$

where

$v = 48.0$ is the final velocity at the moment of impact

$u = 12.0$ is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time it takes for the rock to reach the ground

And solving for t, we find

$t=\frac{v-u}{a}=\frac{48.0-12.0}{9.8}=3.67 s$

6 0

3 years ago

Bonus: What is the velocity of an 8 kg lead shot-put if it has 484 J of energy? Type here to search e

Dima020 [189]

Answer:

11m/s

Explanation:

Given parameters:

Mass of the lead shot = 8kg

Energy of the shot = 484J

Unknown:

Velocity of the shot = ?

Solution:

To solve this problem, we apply the kinetic energy formula;

K.E = $\frac{1}{2}$ m v²

m is the mass

v is the unknown

Now, insert the parameters and solve;

484 = $\frac{1}{2}$ x 8 x v²

484 = 4v²

v² = 121

v = 11m/s

7 0

3 years ago

Particles in a medium through which a sound wave is traveling ​

Particles in a medium through which a sound wave is traveling