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2 years ago

A 2. 0 kg-book is at rest on a 30 o inclined plane. The normal force acting on the book by the inclined plane is most nearly

Physics

1 answer:

Anvisha [2.4K]2 years ago

3 0

The normal force acting on the book by the inclined plane is most nearly 17N.

<h3>What is the normal force?</h3>

When a body is resting on the surface, the reaction force acting perpendicular to the surface of contact and in upward direction is equal to the the weight of the object.

On an inclined plane of 30 degree, a 2 kg block is resting. Its weight mg has two components, mgcos30 and mgsin30.

The Normal force will be equal to

N = mgcos30

N =2 x 9.81 x cos30

N = 17 Newtons

Thus, normal force acting on the book by the inclined plane is most nearly 17N.

Learn more about normal force.

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A sled of mass 8 kg slides along the ice. It has an initial speed of 4 m/s but

marysya [2.9K]

As our story begins, the sled ... whose mass is 8 kg ... is sliding along the ice at a speed of 4 m/s.

The sled's kinetic energy is (1/2 m v²) = (4 kg · 16 m²/s²) = 64 J .

After what seems like only the blink of an eye, the sled is no longer sliding. It is stationary. Motionless. At Rest. Just sitting there !

Its speed has been reduced to zero and ... because kinetic energy is the energy of motion ... the sled's kinetic energy is now also zero. Sixty-four Joules of energy have disappeared !

How can this be ? ! ? We know that energy is conserved. It can never just appear out of nothing, and it can never just disappear into nothing. If energy suddenly appears, it had to come from somewhere, and if energy suddenly disappears, it had to go somewhere. So where did our 64 Joules of kinetic energy go ?

It went into the ice, THAT's where ! We can say that the sled did 64J of work, and melted a thin slick layer of water on the surface of the ice. OR we can say that friction did NEGATIVE 64J of work on the sled, to cancel the 64J that it had originally, sap its kinetic energy, and bring it to rest.

I think <em>choice-B</em> was supposed to say "<em>B. -64J</em>", but somebody typed it sloppily and neglected to proofread it before posting.

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3 years ago

When a 25000-kg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.23cm deeper into the water.

Genrish500 [490]

Answer:

10604 square meters

Explanation:

0.23 cm = 0.0023m

Assume the carrier has a shape of a rectangular box. When the carrier sinks 0.0023m deeper into water, the extra volume submerged into water is the same as the extra water volume being replaced. This extra volume would add an additional buoyant force to counter balance the extra weight created by the 25000 kg fighter.

With g being constant, the mass of the extra water displaced is the same as the mass of the fighter airplane $m_f$ . And mass of the water displaced is its volume V times its density $\rho$

$V\rho = m_f$

$1025V = 25000$

$V = 25000/1025 = 24.39 m^3$

We assume the carrier has a shape of a rectangular box, so this extra displaced volume is the extra depth d = 0.0023 m times cross-section area A

$V = dA$

$24.39 = 0.0023A$

$A = 24.39 / 0.0023 = 10604 m^2$

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3 years ago

What are the properties of elements based on their placement on the periodic table?

Step2247 [10]

The property are on there hope this helps

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3 years ago

Question from my vector statics class (deals with vectors from physics)

jeka57 [31]

Haven't done one like this in awhile but I see no one is answering so I gave it a try. I think it's right but let me know if you see something fishy...

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3 years ago

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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$