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2 years ago

9

An atom that has 117 protons in its nucleus has not yet been made. Once this atom is made, to which group will element 117 belon

g ? Explain ur answer

1 answer:

Pavel [41]2 years ago

3 0

In nomine patris, et filii, et spiritus sancti.

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what times are the moon phases visible? My science teacher said that the phases of the moon are only visible at sertant times of

klio [65]

Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.

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2 years ago

If you drop a 50 gram piece of metal that has a temperature of 110°Celsius into 1000 grams of water at 25°Celsius, what best des

LuckyWell [14K]

If you drop a 50 gram piece of metal that has a temperature of 110°Celsius into 1000 grams of water at 25°Celsius, <span>D.)The water and the metal’s temperature will reach the same temperature. In any system undergoing heat transfer, the objects involved will eventually reach the same temperature, signifying thermal equilibrium.</span>

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2 years ago

Read 2 more answers

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

2 years ago

If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a

polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = $\frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm$

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

3 0

2 years ago

¿que acciones o actitudes de tiempo de inmovilizacion obligatoria descubrimos o hemos fortalecido en casa con familia?

kap26 [50]

Answer:

tenemos más convivencia nos estamos uniendo más y podemos ayudarnos todos mutuamente en cualquier cosa gracias a la inmovilización obligatoria

4 0

2 years ago