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3 years ago

A car goes North a distance of 120 kilometers during a time period of 3.0 hours. What is the average

Physics

1 answer:

DIA [1.3K]3 years ago

8 0

Answer:

the average speed of the car during this time interval is 40 m/s i thinkk..

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A car having an initial velocity of 12 meters per second speed up uniformly to 22 meters per second. This change is made in 4 se

Sonbull [250]

(22-12)/4

10/4

2.5 meters per second sqaured

7 0

3 years ago

Which one of the following is not a fundamental physical quantity??! A. Temperature b. Current c. Area d. Mass

Hatshy [7]

Answer:

Out of this, Area is not a fundamental physical quantity.

3 0

3 years ago

This model shows an example of a fog bank formation. This can happen in the Great Lakes area as warm summer air moves across the

Vadim26 [7]

Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.

<h3>How Fog banks formed?</h3>

Fog banks form at sea where cool air moves quickly over the surface of the ocean that is warm. The cool incoming air lowers the temperature of the air just above the water surface and water vapor condenses into fog.

So we can conclude that Fog bank formation occurs when the moisture in the cooled air condenses, forming a thick fog.

Learn more about fog here: brainly.com/question/18943608

#SPJ1

5 0

2 years ago

400 N of load can be overcome by an effort of 50 N by using a lever. Calculate the mechanical advantage of the lever.

mars1129 [50]

Answer:

load (l)=400N

Effort(E)=50N

mechanical advantage (MA)= load ÷Effort

(ma)=400÷50

(ma)=8

Explanation:

I copy pasted from the answer from the same question. Remember to first check if ur question is there

4 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$