All categories

3 years ago

Heating food under a heat lamp is an example of heat transfer by

Physics

1 answer:

Tcecarenko [31]3 years ago

5 0

<span>Heating food under a heat lamp is an example of heat transfer by <span>Radiation</span></span>

You might be interested in

What is the kinetic energy of a 1 kg pie traveling at a speed of 4 m/s?

Veronika [31]

We have: K.E. = mv² / 2
Here, m = 1 Kg
v = 4 m/s

Substitute their values in the formula,
K.E. = 1×4² / 2
K.E. = 16/ 2
K.E. = 8 J

Finally, answer of your question would be 8 Joules.

Hope this helps!

4 0

3 years ago

Un objeto que tiene como masa 80 kg acelera a razón de 10 m/s2 ¿que fuerza lo impulzo

Gemiola [76]

Answer:

What

Explanation:

7 0

3 years ago

Read 2 more answers

A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that

pav-90 [236]

Answer:

The permittivity of rubber is $\epsilon = 8.703 *10^{-11}$

Explanation:

From the question we are told that

The magnitude of the point charge is $q_1 = 70 \ nC = 70 *10^{-9} \ C$

The diameter of the rubber shell is $d = 32 \ cm = 0.32 \ m$

The Electric field inside the rubber shell is $E = 2500 \ N/ C$

The radius of the rubber is mathematically evaluated as

$r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m$

Generally the electric field for a point is in an insulator(rubber) is mathematically represented as

$E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}$

Where $\epsilon$ is the permittivity of rubber

=> $E * \epsilon * 4 * \pi * r^2 = Q$

=> $\epsilon = \frac{Q}{E * 4 * \pi * r^2}$

substituting values

$\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}$

$\epsilon = 8.703 *10^{-11}$

7 0

3 years ago

Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42º E from the western sta

olchik [2.2K]

Answer:

Explanation:

AB = 110 miles

Let the distance of the western station from fire is d.

As according to the diagram, use Sine law

$\frac{d}{Sin 15}=\frac{110}{Sin 133}$

d = 110 x 0.2588 / 0.73

d = 39 miles

8 0

3 years ago