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lord [1]
4 years ago
15

MATH PHYS PLEASE HELP ME THANK YOU IN ADVANCE

Physics
1 answer:
Troyanec [42]4 years ago
4 0

Explanation:

I don't know what values you have measured, or what <em>Ladybug Revolution</em> is, but I can show you the equations for rotational motion.

Angular displacement is:

Δθ = ω₀ t + ½ αt²

where ω₀ is the initial angular velocity, α is the angular acceleration, and t is time.

Assuming the radius is constant, the arc length is radius times angular displacement.

s = rΔθ

Using the equation for angular displacement, we can see that if two points have the same angular velocity and same angular acceleration, then they also have the same angular displacement.

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The area of a rectangle of length 100
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Answer:

1. 8000cm2

20. kgms2

19. a, b, c

8 0
3 years ago
Find the equivalent resistance of this
MAXImum [283]

Explanation:

1/R = 1/R1 + 1/ R2

1/R = 1/960 + 1/640

1/R = 5 / 1920

<h3> R = 384 ohm </h3>

So , Req = 384 + R3

Req = 384 + 180

<h3> Req = 564 ohm</h3>

\huge\red{A}\pink{N}\orange{S}{W}\blue{E}\green{R}

<h2> Req = 564 ohm</h2>

7 0
2 years ago
Which of the following is the flow of electrons through a wire or a conductor?
Sonja [21]

Explanation:

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7 0
3 years ago
consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti
ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$\frac{Energy}{Area} = \sigma\times T^{4}

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$E= \sigma\times T^{4}

Now -

For Star A →

T_{A} = 4900 K

For Star B →

T_{B} = 9900 K

Therefore -

$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}

\frac{T_{B} }{T_{A} }= 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }

E(B) = E(A) x (\frac{T(B)}{T(A)} )^{4}

E(B) = E(A) x 2^{4}

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

#SPJ4

4 0
2 years ago
What factors do NOT affect friction between two objects? Explain how you know this.
Black_prince [1.1K]

The frictional force between two bodies depends mainly on three factors: (I) the adhesion between body surfaces

(ii) roughness of the surface

(iii) deformation of bodies.

<3

4 0
3 years ago
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