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umka21 [38]
3 years ago
15

En un recipiente de cobre de forma rectangular de aproximadamente 3.5 m largo por 4.5m ancho tiene una temperatura en temporada

de invierno 4 ° C . Calcular su superficie cuando este en verano a una temperatura 40 ° C . ( Cobre 16.6 x 10 ( -6 )
Physics
1 answer:
Darina [25.2K]3 years ago
3 0

Answer:

A_f= 15,769 m²

Explanation:

Este es un ejercicio de dilatación térmica,  

          ΔA = (2α) A₀ ΔT

el arrea de recipiente

          A₀ = L A

          A₀ = 3,5 4,5

          A₀=  15,75 m²

el coeficiente de dilatación térmica es alfa = 16,6 10⁻⁶ C⁻¹

calculemos

            ΔA = 2 16,6 10⁻⁶  15,75 ( 40 -4)

            ΔA = 522,9 (36)  10⁻⁶  

            A= 1,88 10⁻²  m2

el cambio de volumen es

          ΔA = A_f – A₀

          A_f = A₀ +  ΔA

          A_f=  15,75 +1,88 10⁻²  

          A_f= 15,769 m²

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A 1.2kg stone is tied to a string and swung in a vertical circle with a radius of 0.75m. The string can withstand a tension of 4
san4es73 [151]

Hello!

We can begin by summing the forces acting on the stone when it is at the bottom of its trajectory.

Refer to the free-body diagram in the image below for clarification.

We have the force of tension (produced by the string) and the force of gravity acting in opposite directions, so:
\Sigma F = T - F_g

The net force is equivalent to the centripetal force experienced by the stone. Recall the equation for centripetal force for uniform circular motion:
F_c = \frac{mv^2}{r}

m = mass of object (1.2 kg)
v = velocity of object (? m/s)
r = radius of circle (0.75 m)

The centripetal force is the resultant of the forces of tension and gravity, and points upward (same direction as the tension force) since the tension force is greater.

Therefore:
\frac{mv^2}{r} = T - Mg

We can solve the equation for 'v':

mv^2 = r(T - Mg) \\\\v^2 = \frac{r(T - Mg)}{m}\\\\v = \sqrt{\frac{r(T - Mg)}{m}}

Plug in values and solve.

v = \sqrt{\frac{(0.75)(40 - 1.2(9.8))}{1.2}} = \boxed{4.201 \frac{m}{s}}

3 0
2 years ago
A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?
Kazeer [188]
-- She went up for 0.4 sec and down for 0.4 sec.

-- The vertical distance traveled in gravity during ' t ' seconds is

                   D  =  (1/2)  x  (g)  x  (t)²

                       = (1/2) (9.8 m/s²) (0.4 sec)²

                       =    (4.9 m/s²)  x  (0.16 s²)

                       =      0.784 meter        ( B )
4 0
3 years ago
Read 2 more answers
A 75.0 kg man pushes on a 500,000 kg wall for 250 s but it does not move.
TiliK225 [7]

Answer:

he does no work on the wall coz wall didn't move

the wall doesn't move so the energy that he use will be wasted

the force that he put on the wall is also zero since the wall didn't move

8 0
3 years ago
A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou
qwelly [4]

Answer:

-v/2

Explanation:

Given that:

  • a disc of mass m
  • Collides with the wall going through a sliding motion on on the plane smooth surface.
  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

KE_i=\frac{1}{2}. m.v^2

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

KE_f=\frac{1}{4}\times KE_i

KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2

Considering that the mass of the body remains constant before and after collision.

KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

3 0
3 years ago
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