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iVinArrow [24]
3 years ago
13

Describe a situation where your body can act as a lightning conductor.

Physics
1 answer:
Naddika [18.5K]3 years ago
7 0

When someone is struct by lightning, the electricity passes through the body, into the earth. Here, our body acts as a lightning conductor to complete the earthing process.

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

7 0
3 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
discharge occurs when oppositely charged objects get close enough for the air between them to become electrically charged.
Wittaler [7]

Answer:

It does both. Once they get close enough the air does start to get charged, but then they eventually discharge when they touch.

Explanation:

4 0
3 years ago
A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During th
fgiga [73]

Answer:

the final energy of the system is 35.5 kJ.

Explanation:

Given;

initial energy of the system, E₁ = 10 kJ

heat transferred to the system, q₁  30 kJ

Heat lost to the surrounding, q₂ = 5kJ

heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU  is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

E₂ = 10 kJ + 25.5 kJ

E₂ =  35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.

3 0
2 years ago
Which two terms indicate the same area of the body? *
Kruka [31]

Answer:

For example, the toes are anterior to the heel, and the popliteus is posterior to the patella. Superior and inferior, which describe a position above (superior) or below (inferior) another part of the body. For example, the orbits are superior to the oris, and the pelvis is inferior to the abdomen.

Explanation:

4 0
3 years ago
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