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Naily [24]
3 years ago
7

Fizik

Physics
1 answer:
vekshin13 years ago
3 0

Answer:

P = 24.32 W

Explanation:

The question is, "Two resistors 5 Ω and 10 Ω are connected in parallel with a 9 V power supply. Calculate the output power of the power supply.".

The voltage of the power supply, V = 9 V

Resistor 1, R₁ = 5 Ω

R₂ = 10 Ω

The equivalent of parallel combination of resistors is given by :

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_{eq}}=\dfrac{1}{5}+\dfrac{1}{10}\\\\R_{eq}=3.33\ \Omega

The power of the output is given by :

P=\dfrac{V^2}{R_{eq}}\\\\=\dfrac{9^2}{3.33}\\\\P=24.32\ W

So, the output power is equal to 24.32 W.

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The concentration of sodium and potassium ions in the blood and body fluid is regulated by :
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Answer:

Kidney

Explanation:

One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.

Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.

Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.

6 0
3 years ago
A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate
lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air y at time t is given by

y=100\,\mathrm m-\dfrac g2t^2

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-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}

7 0
3 years ago
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

4 0
3 years ago
Question 6 of 10
Anni [7]
The answer is c or b u choose
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3 years ago
A 2.20 kg mass is at the origin. A
Karolina [17]

Answer:

0

Explanation:

F1 = G•2.2•4.66/3² (pointed right)

F2 = G•2.2•4.66/3² (pointed left)

subtract the two to get zero

3 0
2 years ago
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