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Lena [83]
2 years ago
13

How can larger telescopes

Physics
1 answer:
alexira [117]2 years ago
8 0

Answer:

Explanation:

"Larger telescopes" generally means larger apertures. This is the part of the telescope which collects the light. The more light collected, the more clear the picture becomes

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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<u> positive.</u>

<u></u>

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

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

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 <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • 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
2 years ago
At the beginning of a unit on forces, Ms. Alton is leading a class discussion asking her students
VMariaS [17]

Answer:

(iv), (v), (vi) would be incorrect.

Explanation:

(iv) Force isn't transferred from one colliding object to another, but momentum can be.

(v) An object doesn't stop immediately a force stops acting on it. Think of a thrown ball.

(vi) For an object not to move, it means that the net force on the object is zero, and not necessarily that there are no forces acting on the object. For example, an object could be pushed on one side, and be pushed on the other side with an equal force in the opposite direction. The forces would cancel each other and the net force would be zero.

The rest should be correct.

6 0
3 years ago
A 755 N force is used to push a 15 Kg box across the floor. What is the acceleration of the box?
cluponka [151]

Answer:

good question!

Explanation:

3 0
2 years ago
Read 2 more answers
A bus travels east for 3 km the north for 4 km what is its final displacement
lesya [120]
Im about to the math for this right now.
8 0
2 years ago
Two carts, A and B, are connected by a spring and sitting at rest on a track. Cart A has a mass of 0.4 kg and Cart B has a mass
Ilya [14]

Both carts experience the same force but Cart A has a greater speed after the recoil.

The given parameters;

  • <em>Mass of the cart A = 0.4 kg</em>
  • <em>Mass of the cart B = 0.8 kg</em>

Apply the principle of conservation of linear momentum to determine the velocity of the carts after collision;

m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.

F_A = -F_B

Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0
2 years ago
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