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Fed [463]
3 years ago
14

Consider the circuit. Switches are added at points A, B, C, and D. All the switches are closed EXCEPT the switch at position D,

which is left open. What is the result of this?
Physics
2 answers:
lora16 [44]3 years ago
7 0

Answer:

A I just had it on usa test prep

Explanation:

Liono4ka [1.6K]3 years ago
5 0

answer: a) only light 4 will go out.

i did the usatestprep.

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A wire 1.0 m long experiences a magnetic force of 0.50 N due to a
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11.50

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Drag each label to the correct location on the image
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Check and double-click on the red polygon for the Henry Mountains laccolith complex in the folder labeled Problem 1. This comple
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3 years ago
Read 2 more answers
Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to
zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 )  friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5  = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

4 0
3 years ago
Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po
alexgriva [62]

Answer:

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t + 45

Explanation:

1 full revolution is 2\pi. let \theta be the angle of Ron's position.

At t = 0. \theta = 0

one full revolution occurs in 12 sec, so his angle at t time is

\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t

r is radius of circle and it is given as

x = rcos\theta

y = rsin\theta

for r = 30 sec

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t + 45

7 0
2 years ago
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