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defon
3 years ago
6

I need help with this please

Physics
1 answer:
Firdavs [7]3 years ago
6 0
C is the answer to the question
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When a 2.40-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.92 cm.(a) Wh
Anastasy [175]

Answer:

805.48N/m

Explanation:

According to Hookes law

F = Ke

F is the force = mg

F = 2.4×9.8 = 23.52N

e is the extension = 2.92cm = 0.0292m

Force constant K = F/e

K = 23.52/0.0292

K = 805.48N/m

Hence the force constant of the spring is 805.48N/m

3 0
2 years ago
1. Set frequency of wave generator to 5Hz.
ivann1987 [24]

Separate the barriers so they have a 2cm gap between them.

3 0
3 years ago
What are the names of neptunes moons
alexandr402 [8]

Answer:

1. naiad  

2.Thalassa

3 Despina

4 Galatea  

5 Larissa  

6 Hippocamp

7 Proteus ˈ  

8 Triton    

9 Nereid

10 Halimede  

11 Sao

12 Laomedeia

13 Psamathe  

14 Neso

Explanation:

hope this  helped

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
The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F
photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

<em>T</em> - 25 N = (8 kg) (2.4 m/s²)

<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N

5 0
3 years ago
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