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DIA [1.3K]
3 years ago
9

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

lly compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.
Physics
1 answer:
BabaBlast [244]3 years ago
5 0

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

         Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

         Em_f = K = ½ m v²

since friction is zero, energy is conserved

          Em₀ = Em_f

          1 / 2k x² = ½ m v²

          v = \sqrt{ \frac{k}{m} }     x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

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You observe that a negatively charged plastic pen repels a charged piece of magic tape. You then observe that the same piece of
hichkok12 [17]

Answer:

(2) The excess negative charge from the sphere spread out all over your body.

(7) After you touched it, the metal sphere was very nearly neutral.

Explanation:

Plastic pen repels magic tape so magic tape is also  negatively charged . Further , magic tape repels small metal sphere that means small sphere also is negatively charged.

Now when small sphere is touched by a man insulated from ground , the charge is distributed between man and small sphere according to their capacitance .

Since human body will have greater capacitance ,it will acquire larger share of charge . Sphere being of very small size will retain very less charge and it will become almost neutral . Hence it will be   attracted by  charged tape .

5 0
3 years ago
If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links p
Mashcka [7]

The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was

<em>a</em> = ∆<em>v</em> / ∆<em>t</em> = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²

By Newton's second law, the magnitude of the force <em>F</em> is proportional to the acceleration <em>a</em> according to

<em>F</em> = <em>m a</em>

where <em>m</em> is the object's mass. Solving for <em>m</em> gives

<em>m</em> = <em>F</em> / <em>a</em> = (10 N) / (3.52941 m/s²) ≈ 2.8 kg

4 0
2 years ago
Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizon
Tems11 [23]

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

  • Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
  • Fperpendicular = F*cosα =  300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

<u><em>The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.</em></u>

6 0
3 years ago
Starting from rest, a 5.00-kg block slides 2.50 m down a rough 30.0 incline. The coeffi cient of kinetic friction between the bl
UkoKoshka [18]

Answer:

a) W_g =61.25J

b) W_k = -46.25J

c) W_N = 0

d) W_g would be the same.

   W_k would decrease.

   W_N would be the same.

Explanation:

a) On an inclined plane the force of gravity is the sine component of the weight of the block.

F_g = mg\sin(\theta) = 5(9.8)\sin(30^\circ)\\W_g = F_g x = 5(9.8)\sin(30^\circ)2.5 = 61.25J

b) The friction force is equal to the normal force times coefficient of friction.

F_k = -mg\cos(\theta)\mu_k = -5(9.8)cos(30^\circ)0.436 = -18.5 N\\W_k = -F_kx = -46.25J

c) The work done by the normal force is zero, since there is no motion in the direction of the normal force.

d) The relation between the vertical height and the distance on the ramp is

h = x\sin(\theta)

According to this relation, the work done by the gravity wouldn't change, since the force of gravity includes a term of x\sin(\theta).

The work done by the friction force would decrease, because both the cosine term and the distance on the ramp would decline.

The work done by the normal force would still be zero.

5 0
3 years ago
The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul
ozzi
The correct answer is 
B It increases.

In fact, the kinetic energy of a moving object is given by
K= \frac{1}{2}mv^2
where m is the mass of the object and v is its speed. We see that the kinetic energy is proportional to the mass and proportional to the square of the speed: in this problem, the speed of the object remains the same, while its mass increases, therefore the kinetic energy will increase as well.
5 0
3 years ago
Read 2 more answers
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