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3 years ago

Compare and contrast Ursa Major and the Milky Way

1 answer:

7 0

One thing thats the same is there both universes.

One thing diffrent is that in the milky way there is proven life on earth but ursa major is proven without life.

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How far is tau hercules in light years from earth?

Nady [450]

Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.

8 0

4 years ago

If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?

Otrada [13]

Answer: 0.2 hour

Explanation:

Speed = Distance / Time

Time = Distance / Speed

= 12/60

= 0.2 hour

8 0

3 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is positive.

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is positive.

Given:

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

What are the applications of pascal's principle

Murrr4er [49]

Explanation:

The applications are, hydraulic lift- to transmit equal pressure throughout a fluid.
Hydraulic jack- used in the braking system of cars.
use of a straw- to suck fluids, which goes because of air pressure.

<h3>The question simply asks, where pressure can be applied. There are many others, such as lift pump.</h3>

5 0

3 years ago

In physics class, Carrie learns that a force, F, is equal to the mass of an object, m, times its acceleration, a. She writes the

Phantasy [73]

Answer:

2.2 m/s^2

Explanation:

Acceleration = Force / Mass

= 7.92 / 3.6 = 2.2m/s^2

Hope this help you :3

5 0

3 years ago

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