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frosja888 [35]
2 years ago
9

When a wire is made thicker its resistance what?

Physics
1 answer:
NeTakaya2 years ago
7 0
Making a wire thicker has the same effect as making a road wider. It makes it easier for the electron traffic to flow. The resistance decreases, and the current (traffic) increases.
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State the objects in the universe that give out their own light
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All of those are stars.
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3 years ago
A 0.40 kg bead slides on a straight frictionless wire with a velocity of 3.50 cm/s to the right. The
tensa zangetsu [6.8K]

Answer:

Total momentum before collision

P1 =.4 * 3.5 = 1.4       ignoring units here

Total momentum after collision

P2 = .6 * V - .4 * .7 = .6 V - .28

.6 V = 1.4 + .28   momentum before = momentum after

V = 2.8 cm/sec

In 5 sec V moves 2.8 cm/sec * 5 sec = 14 cm

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What is the first law of thermodynamics?​
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it states that energy can neither be created or destroyed

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Read 2 more answers
If you lose control of your vehicle and collide with a fixed object, such as a tree, at 60 m.p.h., the force of impact is the sa
My name is Ann [436]
You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.

First, convert 60 mph to m/s:

60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s

Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.

free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)

H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m

If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.

Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.
7 0
3 years ago
A 15 kg mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done
sashaice [31]
Kinetic energy = (1/2) (mass) x (speed)²

At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules

At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules

The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.

That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
8 0
3 years ago
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