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dalvyx [7]
2 years ago
6

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of -2.0 µC; sphere B carries a charge of -6.0 µC;

and sphere C carries a charge of +4.0 µC. Spheres A and B are touched together and then separated. SpheresB and C are then touched and separated. Does sphere C end up with an excess or a deficiency of electrons and how many electrons is it?Select one:A. deficiency, 6 × 10^13B. excess, 2 × 10^13C. There is no excess or deficiency of electrons.D. excess, 3 × 10^13E. deficiency, 3 × 10^12
Physics
1 answer:
Dmitry_Shevchenko [17]2 years ago
5 0

Answer:

None of the option is correct

A=-4µC, B=0 C, C=0 C

C will be +4.0 µC deficient after the contact

Explanation:

A and B are come in contact together, the charge will flow to establish  equilibrium, and hence becoming: A=-4µC, B=-4µC, C=+4.0 µC

Similarly when C and B touch, the positive and the negative will exact the same force due to equal charge magnitude and become electrically neutral : A=-4µC, B=0 C, C=0 C.

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MissTica
A perpetual motion machine is (as the name implies) a machine that moves perpetually; it never stops. Ever. So if you created one today and set it going, it would keep on going until the Big Freeze<span>. Calling that “a long time” is an understatement of epic proportions</span>
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2 years ago
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Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68
den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, y_0=1500\ m

Initial horizontal velocity is, u_x=688\ m/s

Initial vertical velocity is, u_y=0(\textrm{Horizontal velocity only initially)}

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is S=y-y_0=0-1500=-1500\ m

Acceleration in the vertical direction is due to gravity, g=-9.8\ m/s^2

Therefore, the displacement of the bomb is given as:

S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0
2 years ago
If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat
Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0
2 years ago
A metal bar moves through a magnetic field. the induced charges on the bar are
Dmitry [639]
I would say its a positive cgarge
7 0
2 years ago
You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float
Allushta [10]

Answer:

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

Explanation:

The density of a liquid is inversely proportional to the volume (height) of object submerged in it.

High density liquid possess higher buoyant force preventing objects from submerging.

p ∝ 1/V ∝ 1/h

since V = Ah

pu/pw = hw/hu

pu = pwhw/hu

Where;

p = density

h = height submerged

pu and pw is the density of unknown liquid and water respectively

hu and hw is the height of object submerged in unknown liquid and water respectively

pw = 1000kg/m^3

hu = 4.6cm = 0.046m

hw = 5.8cm = 0.058m

Substituting the given values;

pu = 1000×0.058/0.046

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

5 0
2 years ago
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