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3 years ago

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Jay bought a toy car. To make the car move, he must turn a key attached to a spring. Turning the key winds the spring tight. The

toy car moves when the key is released and the spring unwinds. What’s the main transformation of energy in this example?

2 answers:

lapo4ka [179]3 years ago

5 0

The chemical energy in Jay's body, to kinetic energy in the car

Brilliant_brown [7]3 years ago

4 0

Answer:

Elastic potential energy into kinetic energy

Explanation:

Initially, all the energy is stored in the spring, which is compressed. This energy is called elastic potential energy, and it is equal to

$U=\frac{1}{2}kx^2$

where k is the spring constant and x is the compression of the spring. When the key is turned, the spring is released, and all this energy is converted into kinetic energy, which is the energy due to the motion of the car. The equation that gives this energy is

$K=\frac{1}{2}mv^2$

where m is the mass of the car and v is the speed of the car.

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Fredrick did an investigation on mass and acceleration. He applied a force to a toy car and measured the distance it moved. He t

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Explanation:

B. More mass results in less acceleration.

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3 years ago

Which of the following would be used in luminosity calculations?

Daniel [21]

Answer:

1) joule

2) $kgm^{2}/s^{2}$

3) $10\%$

Explanation:

1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second ( $\frac{J}{s}$ ), which is equal to Watts ( $W$ ).

This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.

Therefore, if we want to calculate luminosity the Joule as a unit will be used.

2) Work $W$ is expressed as force $F$ multiplied by the distane $d$ :

$W=F.d$

Where force has units of $kgm/s^{2}$ and distance units of $m$ .

If we input the units we will have:

$W=(kgm/s^{2})(m)$

$W=kgm^{2}/s^{2}$ This is 1Joule ( $1 J$ ) in the SI system, which is also equal to $1 Nm$

3) The formula to calculate the percent error is:

$\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%$

Where:

$V_{exp}=7.34 (10)^{-11} Nm^{2}/kg^{2}$ is the experimental value

$V_{acc}=6.67 (10)^{-11} Nm^{2}/kg^{2}$ is the accepted value

$\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%$

$\% error=10.04\% \approx 10\%$ This is the percent error

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3 years ago

A beam of light is traveling through a medium at 200,000 km/s. It enters a different medium and speeds up to almost 250,000 km/s

shtirl [24]

Answer

From the question we can see that the in medium 1 speed of light is 200,000 Km/s where as in Medium 2 speed of light is 250,000 km/s.

We can conclude that medium 1 is denser than Medium 2.

In third medium Light speed halts so, the third medium is opaque.

We can say that

Medium 1 can be either water, glass.

Medium 2 will be gas

Medium 3 will be an opaque material.

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3 years ago

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If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if

umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

$P'=P+\rho gh$

Put the value into the formula

$P'=1.01\times10^{5}+1025\times9.8\times10$

$P'=201450\ Pa$

We need to calculate the volume at the surface

Using equation of ideal gas

$PV= RT$

So, for both condition

$PV=P'V'$

Put the value into the formula

$V=\dfrac{201450\times5.5}{1.01\times10^{5}}$

$V=10.97\ L$

Hence, The volume at the surface is 10.97 L.

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olga55 [171]

Answer:0.588..

Explanation:

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