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3 years ago

If humans can't see air but can see water,

Physics

2 answers:

SashulF [63]3 years ago

5 0

Answer:

xd- good question.

Elden [556K]3 years ago

3 0

Answer:

wow, good question my guy.

You might be interested in

A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12

V125BC [204]

Answer:

M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm

a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2 the acceleration

F = M a = 27 gm * 1/3 m/s^2 = 9 dynes net force applied

4 0

2 years ago

A car travels 15 kilometers west in 10 minutes. After reaching the destination, the car travels back to the starting point, agai

jeka94

Speed = (distance traveled) / (time to travel the distance).

Strange as it may seem, 'velocity' is completely different.

Velocity doesn't involve the total distance traveled at all.
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there. So the displacement in driving once around
any closed path is zero, because you end up where you started.

Velocity =

   (displacement during some time)
divided by
      (time for the displacement)

AND the direction from the start-point to the end-point.

For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):

   Speed = (15km + 15km) / (10min + 5min) = (30/15) (km/min)

   = 2 km/min.

Velocity = (end location - start position) / (15 min) = Zero .

5 0

3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

6 0

3 years ago

Guys down to chat with another dude im sus

solniwko [45]

Answer:

lol imposter winsllllllllllllll

5 0

3 years ago

What is it’s vertical distance below the top of the cliff ?

MakcuM [25]

Answer:

It should be 1 meter, but I'm no scientist.

Explanation: IDK

7 0

4 years ago