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3 years ago

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What process is a method of heat transfer but does NOT contribute significantly to heating the surface or atmosphere of the eart

h?
A) conduction
B) convection
C) insulation
D) radiation

1 answer:

klasskru [66]3 years ago

7 0

Answer:

conduction

Explanation:

Conduction is a method of heat transfer, where two objects exchange heat by touching, but does not contribute to moving heat through earth's atmosphere, because the atmosphere is so sparse it does not transmit heat through conduction effectively. Insulation is not a method of heat transfer.

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Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.

Now using Gauss law we have, E be the electric Field at the distance r from the sheet then

$E\times 2A=\dfrac{\sigma A}{\epsilon_0}\\E=\dfrac{\sigma}{2\epsilon_0}$

The Field will be away from the sheet and perpendicular to it.

a) The Electric Field between them

$E_1=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}\\=0$

b)The Electric Field to the right of the sheets

$E_1=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

c)The Electric Field to the left of the sheets

$E_2=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}\\=\dfrac{\sigma}{\epsilon_0}$

3 0

3 years ago

Two drums of the same size and same height are taken.

Gelneren [198K]

Answer:

i) The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) The pressure acting on the base of <em>B</em> will be the same as the pressure acting on the base of <em>A</em>

iii) The pressure on the base of drum <em>A</em> will be slightly less than the pressure on the base of drum <em>B</em>

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

$h_{max}$ = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If <em>A</em> is completely filled, we have $h_A$ = $h_{max}$

Therefore, $P_A$ = $h_{max}$ × $\rho_{liquid}$ ×g

If <em>B</em> is half filled, we have, $h_B$ = (1/2)· $h_{max}$

$P_B$ = (1/2) × $h_{max}$ × $\rho_{liquid}$ ×g

Therefore, $P_B$ = (1/2) × $P_A$

The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) If both <em>A</em> and <em>B</em> are each filled with water (the same liquid), then the pressure on their bases will be $P_A$ = $h_{max}$ × $\rho_{water}$ ×g = $P_B$ , the same, given that the acceleration due to gravity, <em>g</em>, is constant and the same in Nepal and India

iii) If <em>A</em> is filled with water, and <em>B</em> is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

$P_A$ = $h_{max}$ × $\rho_{water}$ ×g < $P_B$ =

The pressure on the base of drum <em>A</em> will be less than the pressure on the base of drum <em>B.</em>

3 0

3 years ago

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed

olasank [31]

Answer:

Average acceleration is $(11.05)g\ m/s^2$

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{v}{t}$

$a=\dfrac{6500\ m/s}{60}$

$a=108.33\ m/s^2$

Since, $g=9.8\ m/s^2$

So, $a=(11.05)g\ m/s^2$

So, the angular acceleration of the missile is $(11.05)g\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

Diano4ka-milaya [45]

Rx= 3.5 km

Ry= 2.9 km

4 0

3 years ago

Read 2 more answers

a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b

SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0

3 years ago