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Rzqust [24]
4 years ago
6

Which of these is a contact force?

Physics
2 answers:
egoroff_w [7]4 years ago
8 0
The correct answer is B.
Vera_Pavlovna [14]4 years ago
7 0
B is the answer because it causes impact and collides together.
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1. acceleration i believe
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3 years ago
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In a RLC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o
Marrrta [24]

Answer:

<h2>Case i) if \omega L > \frac{1}{\omega c}</h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if \omega L < \frac{1}{\omega c}</h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>

Explanation:

As we know that the impedance of the circuit is given as

z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}

when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination

c_{eq} = c_1 + c_2

so it is

c_{eq} = 2c

now we have

z = \sqrt{(\omega L - \frac{1}{2\omega c})^2 + R^2}

Case i) if \omega L > \frac{1}{\omega c}

So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if \omega L < \frac{1}{\omega c}

So initially if the circuit is capacitive in nature then its net impedance will increase after this

7 0
3 years ago
When energy is transferred to air, what happens to the particles of air? (1 point) They move slower. They move faster. They cool
zubka84 [21]
This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster
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Problem 21-40a:
Greeley [361]

Answer:

Part a)

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

q = 3.37 \times 10^{-4} C

Explanation:

As we know that electric force on electric charge is given as

F = qE

here we have

q = 1.6 \times 10^{-19}C

E = 153 N/C

now force is given as

F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N

Gravitational force on electric charge near surface of earth is given as

F_g = mg

F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N

now the ratio of two forces is given as

\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}

\frac{F_e}{F_g} = 2.74 \times 10^{12}

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

F_g = qE

mg = qE

(5.25 \times 10^{-3})(9.81) = q(153)

here we have

q = 3.37 \times 10^{-4} C

8 0
4 years ago
Describe the behavior of magnets
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