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3 years ago

Why do the stars appear like pointed objects

Physics

1 answer:

saw5 [17]3 years ago

8 0

Answer:

because of where we are and our distance away from them, in reality they're giant balls of burning gas, like the sun. some stars are even bigger than the sun

Explanation:

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When two equal forces act on the same object in opposite directions, what is the net force?

sertanlavr [38]

Answer:

Explanation:

Forces with equal magnitudes and opposite directions cancel each other out, so the net force is 0.

6 0

3 years ago

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If the resistance of a circuit increases, that means that the current has to

Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

Current is <em>inversely proportional</em> to the resistance.
from the relation given below, we can clearly see the relation between current and resistance;

V=IR

I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0

3 years ago

shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r

Ainat [17]

Answer:

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0$

$\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(3R-R)}{(R^2+4R^2)^{3/2}}=0$

$\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0$

$\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}$

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3 years ago

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A glass rod is rubbed briefly with silk and the rod becomes able to attract and lift a number of small pieces of paper. If the r

ser-zykov [4K]

Answer:

it will have a stronger attraction force

Explanation:

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3 years ago

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When they are far apart, the momentum of a proton is 3.2 ✕ 10−21, 0, 0 kg · m/s as it approaches another proton that is initiall

Rina8888 [55]

<span>(6.0x10^-22, -1.40x10^-21, 0) kg*m/s Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction. (3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s After the interaction (2.6x10^-21, 1.40x10^-21, 0) kg*m/s and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be 3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of (6.0x10^-22, -1.40x10^-21, 0) kg*m/s</span>

5 0

3 years ago

Why do the stars appear like pointed objects​

Why do the stars appear like pointed objects