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vlada-n [284]
3 years ago
13

When your hand blocks your view of the lamp, where does the shadow of your hand fall?

Physics
1 answer:
LekaFEV [45]3 years ago
8 0

Answer:

in front of you

Explanation:

Your hand blocks the light of lamp , causing a shadow to form in front of you

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Your high-fidelity amplifier has one output for a speaker of resistance 8 Ω. How can you arrange two 8-Ω speakers, one 4-Ω speak
ANTONII [103]

Answer:

(a) 8Ω (b)  Ratio = Parra/P8 ohm = 1

Explanation:

Solution

Recall that,

An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω

Now,

(a) How can  two 8-Ω speakers be  arranged, when one =  4-Ω speaker, and one =12-Ω speaker

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The Lower arm is : 12 ohm, 4 ohm

The Requirement is  = (16 x 16)/(16 + 16) = 8 ohm

(b) compare  your arrangement  power output of with the power output of a single 8-Ω speaker

The Ratio = Parra/P8 ohm = 1

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4 years ago
Who ever answers Frist get marked the Brainly Est
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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
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Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

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W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

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\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

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