All categories

3 years ago

How is it possible you can feel heat from the sun?

Physics

1 answer:

sergey [27]3 years ago

8 0

there are three ways by which we can transfer heat from one place to other

1. Conduction

In this method of heat transfer heat can be transferred from one place to other without actual movement of medium molecules. This is possible in solid conducting mediums

2. Convection

Here heat energy is carried away along with the medium molecules where medium molecules can move from one place to other while heat is transferred. It is possible in liquid and gas but not possible in solids

3. Radiation

In this mode of heat transfer we do not require any medium and here heat transferred by electromagnetic waves

So we can feel heat from sun in form of electromagnetic waves and it is the fastest way to transfer the heat also it do not require any medium

You might be interested in

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

A rock climber wears a 8.1 kg backpack while scaling a cliff. After 28.2 min, the climber is 9.4 m above the starting point. How

mart [117]

Answer:

Explanation:

mass of backpack, m = 8.1 kg

weight of climber, W = 656 N

height raised, h = 9.4 m

time, t = 28.2 min = 28.2 x 60 = 1692 second

weight of backpack, w = m x g = 8.1 x 9.8 = 79.38 N

Work done by the climber on the backpack = mg x h = 79.38 x 9.4 = 746.17 J

Wok done in lifting herself + backpack = (W + w) x h

= (656 + 79.38) x 9.4 = 6912.57 J

Power developed by the climber,P = Total work / time

P = 6912.57 / 1692 = 4.09 W

3 0

4 years ago

A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the

kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

$\theta=tan^{-1}(\frac{v_y}{v_x})$ (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

$v_x=v_o$

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

$v_y^2=v_{oy}^2+2gy$ (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

$v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}$

Finally, you use the equation (1) to find the angle:

$\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°$

The angle below the horizontal is 12.60°

7 0

3 years ago

3 + 2 ∙ 4 = 3 + (2 ∙ 4)

Rzqust [24]

Answer:

11 = 11

Explanation:

6 0

3 years ago

Read 2 more answers

Un atleta tiene en un instante dado una velocidad de 4 m/s. Si a partir de ese instante y durante 2 s adquiere un MRUA con una a

kolbaska11 [484]

Answer:

Velocidad final, V = 8 m/s

Explanation:

Dados los siguientes datos;

Velocidad inicial, u = 4 m/s

Aceleración, a = 2 m/s²

Tiempo, t = 2 segundos

Para encontrar la velocidad final (v), usaríamos la primera ecuación de movimiento;

V = u + at

Sustituyendo en la fórmula, tenemos;

V = 4 + 2*2

V = 4 + 4

Velocidad final, V = 8 m/s

6 0

3 years ago