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densk [106]
3 years ago
12

Jasmine and her partner want to test to see if there is a relationship between how much of her solar panel is exposed to sunligh

t and how fast her solar car will move. She uses a piece of paper to cover the solar panel different amounts of her solar panel. The chart below shows the data she collected.
Which of the following is the best reason for Jasmine and her partner to take three (3) different trials for this experiment?
They had to fill out all of the lines on the chart.
They wanted at least 15 points to put on her graph.
Any one trial might have been done incorrectly.
The car didn’t work at first.

Physics
1 answer:
katen-ka-za [31]3 years ago
7 0
Any one trial might have been done incorrectly.
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A transformer is to be used to provide power for a computer disk drive that needs 6.4 V (rms) instead of the 120 V (rms) from th
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Answer:

The current in the primary is 0.026 A

Explanation:

Using the formula

I1 = (V1/V2)*I2

we have

I1 = (6.4/120)*0.500

I1 = 0.026 A

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Creating laws outlawing discrimination has done little to change prejudiced views or behaviors.
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When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver
NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times m\times(32.3)^2

We need to calculate the escape speed of the planet

Using formula of kinetic energy

\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2

\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2

v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}

v=41.29\ m/s

Hence, The escape speed of the planet is 41.29 m/s.

4 0
3 years ago
100g converted to kg .
natka813 [3]

Answer: 0.1 kilogram

4 0
3 years ago
Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air
Svet_ta [14]

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

Learn more about angular momentum here:

brainly.com/question/7538238

#SPJ10

8 0
2 years ago
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