Which of the following is an example of bad sportsmanship?

prove that the rate of heat production in each of the two resistors connected in parallel are inversely proportional to the resi

Katyanochek1 [597]

I believe that the answer to the question provided above is that with increase in resistance provided with constant current, Power dissipated will be lessen since power loss is high. Low power dissipation has low heat production.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

5 0

3 years ago

The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -41 oC and 900 Pa, respect

Fittoniya [83]

Answer:

Part a)

$\rho = 0.0205 kg/m^3$

Part b)

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

Explanation:

Part a)

As per ideal gas equation we know that

$PM = \rho RT$

here we know that Martian atmosphere is equivalent to that of carbon

so we will have

$P = 900 Pa$

$T = 273 - 41 = 232 K$

now we will have

$(900)(0.044) = \rho (8.31)(232)$

$\rho = 0.0205 kg/m^3$

Part b)

Now for the earth surface the density of air is given for

$P = 101.6 kPa$

$T = 18 ^oC$

so we will have

$PM = \rho RT$

$(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)$

$\rho = 1.22 kg/m^3$

So density of atmosphere at Martian Surface is very less than the density at Earth.

8 0

3 years ago

14. measuring cylinder containing some water stands on a scale pan. A solid ball is lowered into the water. The water level rise

Basile [38]

Answer:

Let d be the density of fluid.

So , Initial reading of balance, F1 =30dg N

After the level reaches 50cm^3

Final reading of balance , F2 =50dg N

Given that difference between final and initial reading is 30g

i.e, F2 −F1

=30 g

⟹50dg−30dg=30g

⟹20dg=30g

⟹d=30g/20g

⟹d=1.5g/cm^3

So, density of fluid is 1.5g/cm^3

8 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

Where is the switch located on this diagram?

Fed [463]

For this case, the switch is located at point B of the diagram.
Remember that point D is the universal symbol for resistance.
In A what you have is a source of power and in C what you have is a cable.
Therefore, the answer for this case is B.

6 0

3 years ago