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3 years ago

Help me please I’ve been stuck on this forever now ☹️

Physics

2 answers:

valina [46]3 years ago

7 0

C, velocity I believe this is the answer to your question

castortr0y [4]3 years ago

5 0

Hey ! If you've already been stuck on it forever, then
a few more days won't make that much difference.

No, I'm just kidding. I wouldn't do that to you.

I'm sure this is all stuff that you've heard before.
Here it is again, all in one place:
____________________________________

-- LOCATION . . . . . a point.
If you move from one point to another point, then you travel a

-- DISTANCE . . . . . the path you followed between two points.
If you're moving right now, then how fast you're moving is your

-- SPEED . . . . . the RATE of how fast your distance is changing.
If we also know what DIRECTION you're moving, then we know your

-- VELOCITY . . . . . your speed AND direction.
If your speed is changing OR your direction is changing, then you have

-- ACCELERATION . . . . . the RATE of how fast your speed or your direction is changing.
____________________________________

Go through it again, in the other direction:

Acceleration . . . the rate of how fast your velocity is changing.

Velocity . . . your speed AND your direction.

Speed . . . the rate of how fast your distance is changing.

Distance . . . the length of the path you walk from one place to another.
___________________________________

For the question on the screen:

Instantaneous acceleration describes how fast a velocity is changing
at a specific instant.

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Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (2 m²), what's t

Vesna [10]

Answer:

196000 N

Explanation:

The following data were obtained from the question:

Height (h) = 10 m

Area (A) = 2 m²

Force (F) =.?

Next, we shall determine the pressure in the tank.

This can be obtained as follow:

P = dgh

Where

P is the pressure.

d is the density of the liquid.

g is acceleration due to gravity

h is the height.

Height (h) = 10 m

Density (d) of water = 1000 kg/m³

Acceleration due to gravity (g) = 9.8 m/s²

Pressure (P) =...?

P = dgh

P = 1000 × 9.8 × 10

P = 98000 N/m²

Therefore, the pressure acting on the tank is 98000 N/m²

Finally, we shall determine the force of gravity acting on the column of water as follow:

Area (A) = 2 m²

Pressure (P) = 98000 N/m²

Force (F) =.?

Pressure (P) = Force (F) /Area (A)

P = F /A

98000 = F/ 2

Cross multiply

F = 98000 × 2

F = 196000 N

Therefore, the force of gravity acting on the column of water is 196000 N

4 0

3 years ago

Read 2 more answers

A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have

Doss [256]

Answer:

$2000\; {\rm cm^{3}}$ .

Explanation:

When the ball is placed in this pool of water, part of the ball would be beneath the surface of the pool. The volume of the water that this ball displaced is equal to the volume of the ball that is beneath the water surface.

The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let $m(\text{ball})$ denote the mass of this ball. Let $m(\text{water})$ denote the mass of water that this ball has displaced.

Let $g$ denote the gravitational field strength. The weight of this ball would be $m(\text{ball}) \, g$ . Likewise, the weight of water displaced would be $m(\text{water})\, g$ .

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

$\text{buoyancy} \ge m(\text{ball})\, g$ .

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

$\text{buoyancy} = m(\text{water}) \, g$ .

Therefore:

$m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g$ .

$m(\text{water}) \ge m(\text{ball})$ .

In other words, the mass of water that this ball displaced should be greater than or equal to the mass of of the ball. Let $\rho(\text{water})$ denote the density of water. The volume of water that this ball should displace would be:

$\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \end{aligned}$ .

Given that $m(\text{ball}) = 2000\; {\rm g}$ while $\rho = 1.00\; {\rm g\cdot cm^{-3}}$ :

$\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})} \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}$ .

In other words, for this ball to stay afloat, at least $2000\; {\rm cm^{3}}$ of the volume of this ball should be under water. Therefore, the volume of this ball should be at least $2000\; {\rm cm^{3}}\!$ .

3 0

2 years ago

Find the angle of depression from the top of a lighthouse 250 feet above water level to the water line of a ship 2.5 miles offsh

Rudik [331]

Height of the lighthouse = 250 feet
Distance of the ship from the shore = 2.5 miles
We already know that
1 mile = 5280 feet
Then
2.5 miles = 5280 * 2.5
= 13200 feet
Now, to find the angle of depression, we can use the formula
θ = tan−1 (13200/250)
 = 1.50 degrees
I hope that this is the answer that you were looking for and the answer has come to your desired help.

4 0

3 years ago

Engineers are starting on a new project to develop technology. Which of these

Ilya [14]

The answer should be C. In this question

3 0

3 years ago

When a bus is moving (especially with a high speed) on the road suddenly stops or suddenly changes its direction, the luggage on

Masja [62]

Answer:

First law of motion

Explanation:

I say this because this example shows how an object is staying persistent unless it's compled to change

(not sure)

7 0

3 years ago