What is the heaviest planet in the worlx

katen-ka-za [31]

Ya it looks great to be in a story

8 0

3 years ago

A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.

Nastasia [14]

<h3>Solution for the above question : -</h3>

Ohm's law states that :

$v = ir$

the terms used are :

$r = resistance$
$v = potential \: \: difference$
$i = current$

let's solve for electric current :

$2 = i \times 40$

$i = \dfrac{2}{40}$

$i = 0.05 \: A$

$i = 50 \: mA$

$\mathfrak{good\: \: luck \: \: for \: \: your \: \: assignment}$

8 0

2 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

3 years ago

6. Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. Which light bulb has the greater filament resi

Nonamiya [84]

$\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}$

P is power
R is resistance

$\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}$

Hence

$\\ \bull\tt\longrightarrow R\propto V$

$\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}$

Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 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

What is the heaviest planet in the worlx​

What is the heaviest planet in the worlx