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antiseptic1488 [7]
3 years ago
7

PLEASE HELP!!! 12-14!!!!!

Physics
1 answer:
kondor19780726 [428]3 years ago
8 0
12-14 equals -2.....
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Which activities demonstrate reaction time the most?
Black_prince [1.1K]

Answer:

A goal keepee catering the ball in time is answer

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3 years ago
Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with
Tanya [424]

Answer:

(a) K_{small}=4839.3J

(b) K_{larger}=17406.4J

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

K=\frac{1}{2}Iw^2

Where I is rotational of inertia of solid cylinder about its central axis.

So

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2

Substitute the given values

So

K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J

For Part (b)

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2

Substitute the given values

K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J

8 0
3 years ago
What is the current in a 10V Circuit if the resistance is 2 ohms
VladimirAG [237]
V=IR can be changed to V/R=I so 10V/2 ohms = 5amps so 5 amps is your answer boss
8 0
3 years ago
You exert of force of 75 newtons on a rock. You push and you push, but you can't budge it. You are exhausted! How much work did
sergiy2304 [10]
You performed 0 work for the fact that work means the distance of movement made on an object not the amount of force it is exposed to. 0 work because it didn't move
7 0
3 years ago
A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o
Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
3 years ago
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