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3 years ago

Identify all graphs that represent motion at

Physics

1 answer:

Vesna [10]3 years ago

3 0

Answer:

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Explanation:

3) For a displacement time graph, straight line denotes constant speed. For a velocity time graph, the graph parallel to time axis denotes constant speed. Hence, the correct option is a) and d).

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Although it shouldn’t have happened, on a dive I fail to watch my SPG and run out of air. If my buddy is close by, my best optio

Airida [17]

Answer:

A) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent

Explanation:

When it is found that you are out of air while under water, first of all don't panic, look for your buddy. If you are unable to do that so, then you need to make an emergency ascent. First try to make a Controlled Emergency Swimming Ascent (CESA). This ascent remains under control and is performed at a safe ascent rate. As you ascend the air in your lungs will expand with decreasing ambient pressure. To avoid an over pressurization injury, always exhale a continuous string of bubbles while going up.

If you are not sure you will make to the surface that leading to inhale the only option is to turn the CESA into a Buoyant Emergency Ascent. To be ready locate your weight system as you ascend. As an addition remove the weight from one of weight pockets and hold it away from your body in preparation of dropping if necessary. Dropping the weight will give you an upward buoyant force which is an uncontrolled buoyant emergency ascent and should be performed only as the last option.

So, according to this, first, always have a look at your SPG. Then, if you are out of air, look for your buddy, if not found then make CESA and the last option will be buoyant emergency ascent.

5 0

3 years ago

You turn on the faucet for your garden hose and let go immediately the end search flying around and spraying water everywhere wh

max2010maxim [7]

It’s Newton’s third law

7 0

3 years ago

A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with unifor

Dahasolnce [82]

Answer:

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is located on the x-axis= $\lambda_1$

Linear charge density of another long line which is located on the y-axis= $\lambda_2$

We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

The direction of electric field at z=a on the positive z-axis is positive z-direction .

Because $\lambda_1$ and $\lambda_2$ are positive and the electric field is applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

6 0

3 years ago

What was also changed in the "Levers" lab when the position of the fulcrum was changed?

erastova [34]

Effort force

Explanation:

When the potion of fulcrum and weight is changed, the mechanical advantage changes.Increasing the distance between the fulcrum and the effort, there is a proportion increase in effort required to lift a load.The ration of the distance from the fulcrum to the position of input and output application gives the mechanical advantage in levers when losses due to friction are not considered.

Learn More

Mechanical advantage in Levers : brainly.com/question/11600677

Keywords : Levers, fulcrum, position

#LearnwithBrainly

4 0

3 years ago

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

3 years ago