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3 years ago

Would a rougher or smoother object absorb thermal energy faster? Why? Give an example.

1 answer:

5 0

Smoother because it will increase energy and when the energy increases it’ll create heat also . Example: A car racing on a smooth road it’ll go faster than a Car speeding on a bumpy and rough road , Hope that helps .

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What are some solutions to global temperature changes?<br>help asap

Nimfa-mama [501]

Answer:

Power your home with renewable energy.

Weatherize, weatherize, weatherize.

Invest in energy-efficient appliances.

Reduce water waste.

Actually eat the food you buy—and make less of it meat.

Explanation:

6 0

3 years ago

A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi

asambeis [7]

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

6 0

3 years ago

Which of the following is an example of increasing friction intentionally? a. waxing skis b. adding grease to gears on a bike c.

Aleonysh [2.5K]

C i think that it tell me if its wrong

7 0

3 years ago

A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

kirza4 [7]

Answer: $v=\sqrt{\frac{FL}{m}}$

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

$F=T=\frac{mv^2}{L}$

$v=\sqrt{\frac{FL}{m}}$

8 0

3 years ago

a 45 kg ice skater initially skating at a velocity of 3 m/s speeds up to a velocity of 5 m/s. calculate the difference in the ma

ad-work [718]

Answer: 90 kgm/s

Explanation:

The momentum (linear momentum) $p$ is given by the following equation:

$p=m.V$

Where:

$m=45 kg$ is the mass of the skater

$V$ is the velocity

In this situation the skater has two values of momentum:

Initial momentum: $p_{1}=m.V_{1}$

Final momentum: $p_{2}=m.V_{2}$

Where:

$V_{1}=3 m/s$

$V_{1}=5 m/s$

So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):

$p=p_{2}-p_{1}=m.V_{2}-m.V_{1}$

$p=m(V_{2}-V_{1})$

$p=45 kg(5 m/s - 3 m/s)$

Finally:

$p=90 kgm/s$

4 0

3 years ago

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