All categories

3 years ago

What is a variable that gives location relative to an origin?

Physics

2 answers:

lora16 [44]3 years ago

5 0

Answer:

The position variable gives location relative to an origin.

Explanation:

Position (displacement) gives location relative to an origin, but distance does not.

AysviL [449]3 years ago

4 0

The variable would be “X”

You might be interested in

A high ____will have a short wavelength.

Sphinxa [80]

Question:

A high ____will have a short wavelength

Answer:

That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength. Light waves have very, very short wavelengths

Explanation:

Hope it help

4 0

3 years ago

Read 2 more answers

Which of the following best defines insulator?

aliya0001 [1]

Answer:

A substance with low ability or no ability to conduct energy.

Such as Rubber,Silicone,Plastic

6 0

3 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

The final speed of the second package is twice as much as the final speed of the first package.

Explanation:

Free Fall Motion

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

3 years ago

The graph shows the progress of a car which traveled 300 km in 5 hrs. Assuming that its speed remained constant, how far had the

lisov135 [29]

If the car travels
300 km.........................5 hrs
?km ..............................2h
to find out we do (300*2)/5=600/5=120km
also
300km/5h=60km/h
in 2 hours
120km

6 0

3 years ago

In order for an object to accelerate, a forcegravityfrictiona swing must be applied.

vodomira [7]

In order for an object to accelerate, a force must be applied. It follows Newton’s second law of motion where it states that a body at rest remains at rest unless a force is acted upon it. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. Keep in mind that when you exert work, you are exerting energy too.

4 0

3 years ago