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olga_2 [115]
3 years ago
8

Help????????????????

Physics
1 answer:
Naddik [55]3 years ago
6 0

Answer:

23.5

Explanation:

Dunno how 2 explain but this is correct 4 sureeeeee.

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When using a spring scale, the measurements you obtain will be in ____.
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The answer is letter c

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(a) Suppose you charge a 2.5 F capacitor with two 1.5 volt batteries. How much charge was on each plate? 7.5 Correct: Your answe
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Answer:

C = Q / V      

Q = C * V = 2.5 F * 3 V = 7.5 Coulombs (assuming the batteries are in series)

N e = Q       where N is the number of electrons and e the electronic charge

N = 7.5 / (1.60E-19) = 4.7E19 electrons

or 4.7 * 10^19 electrons

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The total resistance of the circuit <br>​
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If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm's Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω

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PLEASE HELP!! WILLING TO GIVE 20 POINTS!
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The book is in motion and at the moment there is no one pushing it to the right. (Remember: a force is not needed to keep a moving object moving to the right.) The forces acting upon the book are shown below.

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3 years ago
An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
sweet [91]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

F_{g} = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = F_{g} + W

N = 710 + 240

N = 950 N

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f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

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