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musickatia [10]
3 years ago
6

Which bright object is in shadow

Physics
1 answer:
IRINA_888 [86]3 years ago
5 0
My best guess would be sun because it is bright but is surrounded by shadows on all sides.
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How did the social structure of ancient Rome affect the lives of its people?
Neporo4naja [7]

Answer:

                 

Explanation:

5 0
3 years ago
xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

where

  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

5 0
3 years ago
Opal adds 25 grams of salt to a one-liter glass beaker filled up to its volume mark with pure water. She stirs the water until t
Mars2501 [29]

By what i know i think that the answer would be A a homogeneous mixture.

5 0
3 years ago
The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
3 years ago
Every football field has two 300kg field goal posts separated by 110m of football field (that includes the endzones). What is th
Mama L [17]

Answer:

4.96×10¯¹⁰ N

Explanation:

The following data were obtained from the question:

Mass 1 (M1) = 300 Kg

Mass 2 (M2) = 300 Kg

Separating distance (r) = 110 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Gravitational force (F) =?

The gravitational force between the two goal posts can be obtained as follow:

F = GM1M2 / r²

F = 6.67×10¯¹¹ × 300 × 300 / 110²

F = 6.003×10¯⁶ / 12100

F = 4.96×10¯¹⁰ N

Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N

3 0
2 years ago
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