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2 years ago

8

1. Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fi

sh really is

2 answers:

Nutka1998 [239]2 years ago

8 0

Answer:

Closer.

Explanation:

Due to the refraction of light passing through water and air.

pychu [463]2 years ago

3 0

Answer: I believe farther. Correct me if I am wrong

Explanation:

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MAXImum [283]

Answer:

true

Explanation:

Newton is the measure of the force with turns to be gravity multiplying the mass. Thus, the forces acts on the particles in the direction of the movement of the particles

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3 years ago

You drop a rock from rest out of a window on the top floor of a building, 20.0 m above the ground. When the rock has fallen 5.00

Andrej [43]

Answer:

You drop a rock from rest out of a window on the top floor of a building, 30.0 m above the ground. When the rock has fallen 3.00 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the ...... rest out of a window on the top floor of a building, 30.0m above the ground. ... You Notice That Both Rocks Reach The Ground At The Exact Same Time. ... You drop a rock from rest out of a window on the top floor of a building, 30.0m ... When the rock has fallen 3.20 m, your friend throws a second rock straight down from ...

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2 years ago

How are mountain ecosystems different from other ecosystems?

Andreyy89

Answer:

They host three or more distinct ecosystems at different elevations.

Explanation:

I checked it, and its right.

3 0

2 years ago

Can a body have zero velocity and finite acceleration?Explain

sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0

3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(c) a = 2874.28m/s^2

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago