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2 years ago

Why is solar energy transferred to earth by radiation

Physics

1 answer:

kodGreya [7K]2 years ago

4 0

Answer: Because between the Earth and Sun there is vacuum

There are three ways in which heat is transmitted:

1. By Conduction, when the transmission is by the direct contact.

2. By Convection, heat transfer in fluids (like water or the air, for example).

3. By Radiation, by the electromagnetic waves (they can travel through any medium and in vacumm) .

So, the space between the Earth and the Sun is vacuum, this means the energy cannot be transmitted by convection, nor conduction. It must be transmitted by electromagnetic waves that are able to travel with or without a medium, and this is called radiation.

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

2 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

2 years ago

1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?

Alik [6]

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

3 0

2 years ago

What is the monthly cost of using a 750 W refrigerator that runs for 8 hours every day, if the cost per kWh is $0.23?

disa [49]

Answer:

Assuming there are 28 days in each month,

750W = 0.75kW

Cost of electric bill = 0.75 × 8 × 28 × $0.23

= $38.64

8 0

1 year ago

An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?

r-ruslan [8.4K]

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

$\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%$

so, the efficiency of heat engine is 13.7%.

5 0

2 years ago