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4 years ago

I THINK I KNOW THE ANSWER BUT I NEED VERIFICATION!!!!!!!

2 answers:

4 0

Yes, the answer is in fact C. All of the above.

Variable stars are stars that change in brightness. The change could be due to the physical change in the star or it could also be caused when another star crosses another, or in other words eclipse.

Umnica [9.8K]4 years ago

4 0

The answer is All of the above. Variable Stars include White Dwarf Stars, Red Giant Stars & Binary Stars. A white dwarf is what stars like the Sun become after they have exhausted their nuclear fuel. A red giant star is a dying star in the last stages of stellar evolution. Binary stars are two stars orbiting a common center of mass. The brighter star is officially classified as the primary star, while the dimmer of the two is the secondary .

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How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

4 years ago

Please show steps as to how to solve this problem Thank you!

liubo4ka [24]

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1 * X2 =

62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y will be out of the page

r X F for F1 will be into the page so the torque must be negative

6 0

3 years ago

resistance is inversely proportional to cross sectional area; is it same as cross sectional area inversely proportional to resis

mart [117]

The greater the cross sectional area of the condoctor, the greater the number of electrons that move and contribute to the current. Having a larger current for the same voltage means having a larger conductance. Since resistance is the inverse of conductance, cross sectional area is inversely related to the resistance.

3 0

3 years ago

It says find the slope for each line I'm stuck on number one can you help me

Allushta [10]