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alukav5142 [94]

2 years ago

14

A wave front has the form of a

1 answer:

podryga [215]2 years ago

3 0

Your answer is "<span>surface of a sphere"

Hope this helps.</span>

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Quarks are thought to be the basic component of what?

vladimir2022 [97]

<span>Quarks are thought to be the basic component of protons and newtons.</span>

5 0

2 years ago

Read 2 more answers

I don't know how to figure it out

drek231 [11]

Nice couch lol
and aluminum I think

4 0

3 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

White raven [17]

Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

4 0

3 years ago

A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an

Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0

2 years ago

Read 2 more answers

In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of

nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

$F_g = F_c$

$\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}$

now we can say that kinetic energy of satellite is given as

$KE = \frac{1}{2}M_s v^2$

$KE = \frac{GM_sM_p}{2r}$

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

$U = -\frac{GM_sM_p}{r}$

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0

3 years ago