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9966 [12]
4 years ago
8

A horse pulls forward on a carriage with a given force. By Newton's Third Law, the carriage must be pulling on the horse backwar

d with an equal and opposite force. Given this, what explains why the horse and carriage can move forward? Two brown horses pulling wooden carriage Image from Wikimedia Commons, CC BY-SA 2.0 Choose 1 answer: Choose 1 answer: A The cart is rolling on wheels while the horse's hooves have traction with the ground B The forward force of the horse is just big enough to overcome the backward force of the cart and start the cart forward C The cart's force is only in reaction to the horse's force so it does not define direction of movement D There is a brief moment where the horse pulls before the reaction force kicks in E The forward and backward forces are equal, so it actually can't move forward
Physics
2 answers:
vladimir1956 [14]4 years ago
7 0
Look at Newton's second law. It says if the forces on ONE object are balanced then the object can't accelerate.

The horse pulls on the carriage and the carriage pulls on the horse. The action force and the reaction force act on DIFFERENT objects.
WITCHER [35]4 years ago
5 0
The horse is rolling on wheels while the horse's hooves have traction to the ground.
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Light with a photon energy of 3 ev impinges on the surface of a material with work function 0.6 ev to eject electrons. what is t
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Answer:

KE = incident energy - work function

The maximum KE will be (3 - .6) eV = 2.4 eV

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2 years ago
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
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Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

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