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worty [1.4K]
2 years ago
12

I''l give brainliest. Please help. I hold 2 objects about 0.1 meters apart. What is the electrostatic force between the two obje

cts if each has a charge of 4.6 x 10^-16 C?
Physics
1 answer:
matrenka [14]2 years ago
7 0

Answer:

F = K Q1 Q2 / R^2       where K = 9 * 10E9  (1 / 4 pi ∈0)

F = 9.00E9 * (4.6E-16)^2 / .01 = 1.90E-19 N

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How much work, in N*m, is done when a 10.0 N force moves an object 2.5 m?
Alik [6]
W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
8 0
3 years ago
If ?h°rxn and ?s°rxn are both negative values, what drives the spontaneous reaction and in what direction at standard conditions
irakobra [83]

The release of free energy drives the spontaneous reaction.

Spontaneity can be <span>determined using the change in </span>Gibbs free energy (the thermodynamic potencial):

delta G=delta H – T*delta S

where delta H is the enthalpy and delta S is the entropy.

The direction (the sign) of delta G depends of the changes of enthalpy and entropy. If delta G is negative then the process is spontaneous.

In our case, both delta H and delta S are negative values, the process as said is spontaneous which means that it may proceed in the forward direction.

6 0
3 years ago
How long does it take long a light from the sun to reach the earth if it travels a distance of 1.5 × 10^11m? (velocity of light
mojhsa [17]

Answer:

Therefore, light travelling at 3.0x10^8 meters per second takes 500 seconds (8 minutes, 20 seconds) to reach the Earth, which is 1.5x10^11 meters away from the sun

Explanation:

7 0
3 years ago
Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
3 years ago
Convert the speed of light 3.0x10^8 m/s to km/day
Aleksandr [31]

Answer: 2.592 \times 10^{8}km/day

1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day



7 0
3 years ago
Read 2 more answers
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