A rock is dropped from rest and falls 50 m determine the velocity upon hitting the groung?

What shape is the graph produced by a force vs acceleration graph?

hammer [34]

The answer to the question<u> What shape is the graph produced by a force vs acceleration graph</u> is A. Linear

Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.

Since this is a linear relationship, the graph of force vs acceleration will be linear.

The answer to the question<u> What shape is the graph produced by a force vs acceleration graph</u> is A. Linear

Learn more about graphs here:

brainly.com/question/24322515

6 0

2 years ago

A diver is is pushed upwards by a diving board. She weighs 739 N and accelerates from rest to a speed of 4.60 m/s while moving 0

gavmur [86]

Answer:

Answer A

Explanation:

It seems more likely

8 0

3 years ago

4. How do ordinary objects behave in microgravity?

Nikolay [14]

Answer:

Bubbles paused

Explanation:

the air bubble doesn't rise because it is no lighter than the water around it—there's no buoyancy. The droplet doesn't fall from the leaf because there's no force to pull it off. It's stuck there by molecular adhesion.

for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth

On Earth, the buoyancy of the air bubbles causes them to rise to the top together, creating a segregation between air and water. However, in microgravity, nothing forces the air bubbles to interact and thus rise together, Green said.

8 0

3 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$