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antoniya [11.8K]
3 years ago
14

A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF. What is the amount of charge stored i

n this capacitor?
Physics
1 answer:
ioda3 years ago
8 0
C = 4 \ \mu F = 4 \cdot 10^{-6} \ F. \newline
q = Cu = 4 \cdot 10^{-6} \cdot 2 = 8 \cdot 10^{-6} = 0.000008 \ C.
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What is a fact about incline planes?
SVETLANKA909090 [29]
They are incline hope this helps!
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2 years ago
A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m
erastovalidia [21]

Answer:

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = k_{f} -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W -W_{fr} = k_{f} -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

 

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

8 0
3 years ago
Briefly describe the formation of the planets from the solar nebula
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<span>The formation of the Solar System began 4.6 billion years ago with the gravitational collapse of a small part of a giant molecular cloud. Most of the collapsing mass collected in the center, forming the Sun, while the rest flattened into a proto-planetary disk out of which the planets, moons, asteroids, and other small Solar System bodies formed.


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6 0
3 years ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is
sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

5 0
3 years ago
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