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charle [14.2K]
3 years ago
6

Có hai bình chứa được nối với nhau bằng một ống nhỏ có thể tích không đáng kể. Một

Physics
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

a b c d e f g h I love you will you marry me

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(will give brainliest to the correct person!)
Neporo4naja [7]

Answer:

Explanation:

When a force hits something, an equal amount of force is exerted back on it.

3 0
3 years ago
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Which of these systems is an oscillator? A. A barrel rolling down the hill B. A child sitting on a swing a skater falling on the
Paladinen [302]

Answer:

Option B:

A child sitting on a swing.

Explanation:

When we hear the word oscillator, a good example is the pendulum bob of a grandfather clock. We can picture the motion to get a perfect understanding of its path of motion and relate it to other systems of motion in our everyday life.

An oscillator is a system that moves in such a way that it reverses its direction after a period of time. It can be seen as a "to-and-fro" motion.

From the options, a child sitting on a swing is the perfect example of an oscillating system because the child will be moving forwards and backwards, alternately reversing the direction of motion with time.

7 0
3 years ago
What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?
Lisa [10]
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x² x = 0.0076 m ↠(a)</span>
7 0
3 years ago
The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge
nataly862011 [7]

Answer:

(a) surface area of the plate will be equal to 1.129m^2

(b) Charge on the capacitor is equal to 1.5\times 10^{-6}C

Explanation:

We have given spacing between the plates d = 0.05 mm = 5\times 10^{-5}m

Value of capacitance C=1\mu F=10^{-6}F

(A) Capacitance of a parallel plate capacitor is equal to C=\frac{\epsilon _0A}{d}

So 10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}

A=1.129m^2

So surface area of the plate will be equal to 1.129m^2

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to Q=CV

So Q=1.5\times 10^{-6}C

5 0
3 years ago
Hi i need help pls help me
r-ruslan [8.4K]
The correct answer would be the last one
4 0
3 years ago
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