Ask question

Ask question

All categories

3 years ago

13

In the Earth's mantle, heat is transferred in large convection currents. Within these currents,

1 answer:

Inessa05 [86]3 years ago

4 0

In the Earth's mantle, heat is transferred in large convection currents. Within these currents "hotter rock rises and cooler rock sinks".

Option B

<u>Explanation:</u>

Heat rises occur and the hotter rock gets farther away from the Earth's overheated core it starts to chill out and continues to sink back towards the Earth's overheated center where it gets reheated and also the entire procedure begins again. It's what a current in convection is.

Convection currents are part of what keeps the Earth's atmosphere through global circulation. The convection currents contribute to conditions in the air and sea. Magma travels in convection currents at Earth's mantle. The hot cores heat the above substance, allowing it to rise to the crust where it cools.

You might be interested in

Why a switch is connected in phase wire and never is neutral wire?

Naddika [18.5K]

If you had the wire connected to a neutral wire, you would never get a charge. You could receive a charge pulse from the phase wire.

3 0

2 years ago

Read 2 more answers

3.0E-17 joules divided by 1.0E -27 kgm/s equals?

insens350 [35]

Here we have two physical quantities which we have to divide

$P_1 = 3.0 \times 10^{-17} J$

other is

$P_2 = 1.0 \times 10^{-27} kg m/s$

now we need to find its division

$\frac{P_1}{P_2} = \frac{3.0\times 10^{-17}}{1.0\times 10^{-27}}$

now we have

$\frac{P_1}{P_2} = 3.0\times 10^{10} Js/kgm$

7 0

2 years ago

Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy

Semenov [28]

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

Q_lost = - Q_absorbed

The latent heat is

Q_absorbed = m L

The heat given by the body

Q_lost = M $c_{e}$ ΔT

where m is the mass of sweat and M is the mass of the body

m L = M c_{e} ΔT

m = M c_{e} ΔT / L

let's replace

m = 90 3.500 1.8 / 2.42 10⁶

m = 0.2343 kg

reduced to grams

m = 0.2342 kg (1000g / 1kg)

m = 23.42 g

the correct answer is c) 23 g

8 0

2 years ago

Please help, thank you,

Elena-2011 [213]

same, stapler, gravity, motion, acceleration

6 0

2 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago