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3 years ago

10 PTS! HELP!

1 answer:

6 0

Shortsightedness is corrected using a concave (curved inwards) lens which is placed in front of a myopic eye, moving the image back to the retina and making it clearer. Longsightedness is corrected using a convex Hyperopia is known to you probably as farsightedness(outward facing) lens.

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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) = 285 joules.

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

2 years ago

What is the best piece of advice you have ever received or given?

Afina-wow [57]

Answer:Your life is your responsibility.

Explanation:

4 0

2 years ago

Read 2 more answers

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

7 0

3 years ago

If an isotope has 8 protons and 10 neutrons in it's nucleus, what is it's atomic mass?

viva [34]

The atomic mass is always equal to the sum of protons and neutrons in the nucleus. If you add the number of protons and neutrons (8 + 10) = 18 you will find that the atomic mass is 18.

8 0

3 years ago

A particle moves in a straight line so that time t seconds its displacement is x metres from a fixed point is given by x=3t^2 +2

Andru [333]

Answer:

-5 meters

Explanation:

x = 3t² + 2t − 5

The initial displacement is when t = 0.

x = 3(0)² + 2(0) − 5

x = -5

6 0

3 years ago