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galina1969 [7]
3 years ago
13

Which statement accurately describes the outer planets?

Physics
2 answers:
tiny-mole [99]3 years ago
6 0
The outer planets have a high gravity due to their large size
True [87]3 years ago
5 0

Answer:

Its D on Edge 2020 just took the quiz

Explanation:

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Um ladrão tenta fugir sozinho carregando em suas mãos uma mala cheia de barras de ouro. A densidade do ouro é igual a 20 g/cm³,
Roman55 [17]
I believe the answer to your question is A
4 0
3 years ago
Three resistors of 10.0 W, 20.0 W, and 25.0 W are connected in parallel across a 100-V battery. What is the equivalent resistanc
leva [86]
In a parallel connection, the equivalent resistance is the summation of the inverse of each individual resistances. It is mathematically expressed as 1/ Req = 1/10 +1/20 + 1/25 = 5.263 ohms. Also, the voltage across each resistor is equal to the input voltage, therefore I = 100 / 10 = 10 Amps. I hope this helped you.
8 0
3 years ago
I WILL GIVE BRAINLIEST....<br>Determine the value of F...​
aleksley [76]
F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.

Or it could be 15 if both top and bottom are 25. 10+15= 25.
6 0
3 years ago
A garrafa térmica (também conhecida como "vaso de Dewar") é um dispositivo extremamente útil para conservar, no seu interior, co
igor_vitrenko [27]

Answer:

A opção A está correta.

O sistema formado pela garrafa térmica e a água perde 400 cal de calor para o meio ambiente.

Option A is correct.

The system formed by the thermos and the water loses 400 cal of heat to the environment.

Explanation:

Quando a temperatura de um sistema reduz, fica claro que o sistema perdeu calor ou energia térmica. Como a temperatura é um dos indicadores mais claros disso, esta conclusão é hermética e correta.

Mas, para saber a quantidade de calor perdida para o meio ambiente, agora fazemos alguns cálculos de energia térmica.

Transferência de calor de ou para o sistema de água e garrafa térmica = c × ΔT

c = capacidade térmica do sistema de água e garrafa térmica = 80 cal /°C

ΔT = Alteração da temperatura do sistema de água e garrafa térmica = (temperatura final) - (temperatura inicial) = 55 - 60 = -5°C

Calor transferido = 80 × -5 = -400 cal.

O sinal de menos mostra que o calor é transferido para fora do sistema, ou seja, o calor é perdido no sistema.

Espero que isto ajude!!!

English Translation

The thermos (also known as "Dewar vase") is an extremely useful device to conserve bodies (essentially liquid) at high temperatures, minimizing energy exchanges with the environment, which is generally colder. A thermos contains water at 60 o C. The thermos + water set has a thermal capacity of C = 80 cal / o C. The system is placed on a table and, after a considerable period of time, its temperature decreases to 55 o C. In this case, it is concluded that the system formed by the thermos and the water inside:

a) lost 400 cal. B) gained 404cal. C) lost 4 850 cal. D) gained 4 850 cal. E) did not exchange heat with the external environment.

Solution

When a system's temperature reduces, it is clear to conclude that the system has lost heat or thermal energy. Since temperature is one of clearest indicators of this, this conclusion is airtight and correct.

But, to know the amount of heat lost to the environment, we now do some thermal energy calculations.

Heat transferrred from or to the water and thermos system = c × ΔT

c = heat capacity of the water and thermos system = 80 cal/°C

ΔT = Change in temperature of the water and thermos system = (final temperature) - (initial temperature)

= 55 - 60 = -5°C

Heat transferred = 80 × -5 = -400 cal.

The minus sign shows that the heat is transferred out of the system, that is, the heat is lost from the system.

Hope this Helps!!!

7 0
3 years ago
A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel
marusya05 [52]

(a) 7.32\cdot 10^7 Hz

The frequency of an electromagnetic waves is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

\lambda=4.1 m is the wavelength of the wave in the problem

Substituting into the equation, we find

f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz

(b) 4.60\cdot 10^8 rad/s

The angular frequency of a wave is given by

\omega = 2\pi f

where

f is the frequency

For this wave,

f=7.32\cdot 10^7 Hz

So the angular frequency is

\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s

(c) 1.53 m^{-1}

The angular wave number of a wave is given by

k=\frac{2\pi}{\lambda}

where

\lambda is the wavelength of the wave

For this wave, we have

\lambda=4.1 m

so the angular wave number is

k=\frac{2\pi}{4.1 m}=1.53 m^{-1}

(d) 1.03\cdot 10^{-6}T

For an electromagnetic wave,

E=cB

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

I=\frac{E^2}{2\mu_0 c}

where we have

E = 310 V/m is the amplitude of the electric field

\mu_0 is the vacuum permeability

c is the speed of light

Substituting into the formula,

I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2

(g) 1.53\cdot 10^{-8} kg m/s

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

\frac{dp}{dt}=\frac{A}{c}

where the <S> is the magnitude of the Poynting vector, given by

=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2

and where the surface is

A = 1.8 m^2

Substituting, we find

\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s

(h) 8.47\cdot 10^{-7} N/m^2

For a surface that totally absorbs the wave, the radiation pressure is given by

p=\frac{}{c}

where we have

=254.2 W/m^2

c=3\cdot 10^8 m/s

Substituting, we find

p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2

8 0
3 years ago
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