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2 years ago

11

If the sun is a medium sized star, why does it look bigger than others?

2 answers:

hjlf2 years ago

8 0

Answer:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller. The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Explanation:

nata0808 [166]2 years ago

5 0

Answer:

The sun looks bigger than other stars because it is closer to the Earth, distance makes it look larger

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How efficient is a pulley system if it enables you to lift a 600.0 Newton engine 0.600 meters if you exerted 35.7 Newtons of for

muminat

Answer:

η = 0.882 = 88.2 %

Explanation:

The efficiency of the pulley system can be given as follows:

$\eta = \frac{W_{out}}{W_{In}}\\\\$

where,

η = efficiency of pulley system = ?

W_out = Output Work = (600 N)(0.6 m) = 360 J

W_in = Input Work = (35.7 N)(11.43 m) = 408.051 J

Therefore,

$\eta = \frac{360\ J}{408.051\ J}$

<u>η = 0.882 = 88.2 %</u>

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2 years ago

The Stronger the force that is exerted on an object,the ____ the change in velocity will be.

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Which of the following words BEST describes scientists’ current knowledge about the universe?

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Answer:

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3 years ago

Three wires meet at a junction. wire 1 has a current of 0.40 aa into the junction. the current of wire 2 is 0.73 aa out of the j

Mnenie [13.5K]

The magnitude of the current in wire 3 is (I₃)= 0.33A

<h3>How to calculate the value of the magnitude of the current in wire 3 ?</h3>

To calculate the magnitude of the current in wire 3 we are using the Kirchhoff’s current law,

I₁ + I₂ + I₃ = 0

Where we are given,

I₁ = current in wire 1

=0.40 A.

I₂ = current in wire 2

= -0.73 A.

We have to calculate the magnitude of the current in wire 3, I₃

Now we put the known values in above equation, we get,

I₁ + I₂ + I₃ = 0

Or, I₃ = -.(I₁ + I₂)

Or, I₃ = -.(0.40 - 0.73)

Or, I₃ = 0.33 A

From the above calculation, we can conclude that the current in wire 3 is I₃ = 0.33 A

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1 year ago

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

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Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

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2 years ago

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