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nevsk [136]
3 years ago
14

El cuerpo de una persona está cubierto con 1.6 m2 de ropa de lana. El grosor de la lana es 2.0 x 10-3 metro. La temperatura en l

a superficie exterior de la lana es 11 C, y la temperatura de la piel es de 36 ˚C. klana=0.04 W/m ˚C. ¿Cuánto calor por segundo pierde la persona debido a la conducción?. R/800 J/s.
Physics
1 answer:
netineya [11]3 years ago
8 0

Answer:

<h2>Rate of heat transfer from the body of the person is given as</h2><h2>\frac{dQ}{dt} = 800 J/s</h2>

Explanation:

As we know that the rate of heat transfer due to thermal conduction is given as

\frac{dQ}{dt} = \frac{kA(T_2 - T_1)}{x}

so we have

k = 0.04 W/m C

A = 1.6 m^2

x = 2\times 10^{-3} m

\frac{dQ}{dt} = \frac{(0.04)(1.6)(36 - 11)}{2 \times 10^{-3}}

\frac{dQ}{dt} = 800 J/s

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A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10
Charra [1.4K]

Answer:

u(t)=1.15 \sin (8.68t)cm

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

Where g=980 cm/s^2

u(t)=Acos8.68 t+Bsin 8.68t

u(0)=0

Substitute the value

A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

Substitute u'(0)=10

8.68B=10

B=\frac{10}{8.68}=1.15

Substitute the values

u(t)=1.15 \sin (8.68t)cm

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0
3 years ago
1. You serve a volleyball with a mass of 2.1kg. The ball leaves your
Darina [25.2K]

The kinetic energy is 945 joules.

Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.

The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.

The mass of the volleyball is 2.1 kg.

The speed of the ball when the ball leaves the hand is 30 m/s.

m = 2.1 kg

v = 30 m/s

The kinetic energy of an object is given as:

KE = (1/2 ) × m × v²

KE = (1 / 2) × 2.1 kg × ( 30 m/s)²

KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s

KE = 2.1 kg × 15 m/s × 30 m/s

KE = 945 J

Learn more about kinetic energy here:

brainly.com/question/8101588

#SPJ9

6 0
1 year ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia
Bingel [31]

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

or

v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

5 0
3 years ago
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
Darya [45]

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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