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2 years ago

Please help me, this is a physics test.

Physics

1 answer:

sweet-ann [11.9K]2 years ago

4 0

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

$v_{f}=v_{o}+a*t\\$

where:

Vf = final velocity = 11 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 3 [s]

$11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]$

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Two point charges are placed on the x axis as follows: charge q1=+3.75nc is located at x=0.205m and charge q2=−5.60 nc is at x=+

Mademuasel [1]

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3 years ago

A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delive

NARA [144]

Answer:

90 hp

Explanation:

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3 years ago

How would improvement in use of renewable energy sources impact climate change sea-level rise?

bonufazy [111]

Answer:

Almost immeasurably small.

Explanation:

The STORY is that humans are BAD for the environment and have caused a HUGE change in the amount of CO₂ in the atmosphere.

Let's look at the reports and draw our own conclusions.

Current CO₂ levels are 409.8 parts per million (PPM)

at the beginning of the Industrial revolution in the 1700's, the presumed beginning of the huge increase in CO₂ the level was about 280 PPM

For perspective lets assume we capture the whole atmosphere and squish it down to 2400 one liter bottles of air

That's 100 cases of 24 bottles per case.

We now separate all the air components into their own bottles

Nitrogen is 78% of our air, so we subtract 78 cases from our 100 leaving 22

Subtracting Oxygen at 21% of air leaves 1 case of liter bottles left

Of those 24 bottles, Argon makes up 0.93% of air so we subtract 22 bottles

The remaining two bottles contain all of the other gasses in our air, One of those bottles contains CO₂.

If we take the CO₂ levels from the 1700's at about 280 PPM as a baseline and assume ALL of the increase is human caused, that is (410 -280) / 280 = 46 % of the total.

The human caused addition of CO₂ would be 460 mililiters out of 2400 liters over the course of 250 years

The claim is, that less than half of a liter of CO₂ out of 2400 liters of air is responsible for heating not only the gas in all the other bottles but also the surface of the earth itself.

Personally, it boggles my mind.

And it says NOTHING of a far more powerful greenhouse gas that is far more prevalent in the atmosphere...water vapor.

Water vapor is about 1% of air at sea level and about 0.4% overall. It was not considered in the above analysis because water vapor can condense out and is not a constant in the air.

Notice that there is about 100 times the amount of water vapor in the air as compared to CO₂. Water vapor also has between 4 and 8 times the greenhouse effect that CO₂ does.

Makes one wonder why we choose to pick on CO₂.

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3 years ago

1 example of a conductor and 1 example of a insulator in your EVERYDAY world.

ratelena [41]

Answer: Examples of conductors include metals, aqueous solutions of salts (i.e., ionic compounds dissolved in water), graphite, and the human body. Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.

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2 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

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3 years ago