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telo118 [61]
3 years ago
14

I WILL GIVE BRAINLIEST TO THE CORRECT ANSWER! Fill in the blanks

Physics
1 answer:
ZanzabumX [31]3 years ago
4 0

Answer:

See the explanation below.

Explanation:

<u>meters</u> per <u>second</u> per <u>second</u> = (m/s/s) = (m/s^2)

This means that the body increases or decreases its speed 1 (m/s), for every second that passes. An easy example to understand is the acceleration of gravity g = 10 [m/s^2]. This means that if a body is thrown from a 100 (m) high tower, for every second it passes the speed will increase by 10 [m/s].

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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
What do you mean by resistance of conductor?state it’s unit.
dalvyx [7]

Answer:

its unit is Ohm

Explanation:

Resistance means material which resist the passing current  through it and the value of resistance says how much the material is resisting the current and it temperature dependent and the unit is Ohm.

7 0
3 years ago
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine the
valentinak56 [21]

Answer:

ma = 48.48kg

Explanation:

To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:

T=2\pi\sqrt{\frac{m_c}{k}}     (1)

mc: mass of the chair

k: spring constant = 600N/m

T: period of oscillation of the chair = 0.9s

You solve the equation (1) for mc, and then you replace the values of the other parameters:

m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg    (2)

Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:

T': period of chair when the astronaut is sitting = 2.0s

M: mass of the astronaut plus mass of the chair = ?

T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg (3)

Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :

m_a=M-m_c=60.79kg-12.31kg=48.48kg

The mass of the astronaut is 48.48 kg

3 0
3 years ago
Where are alkaline earth metals found on the periodic table?
mojhsa [17]

Answer:

B

Explanation:

8 0
2 years ago
True or false. The planet's speed never changes. That is why we have a stable orbit.
Svetradugi [14.3K]
This answer is true the earth always stays at one speed
8 0
2 years ago
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