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4 years ago

How many electrons would it take to equal the mass of a proton? Use two significant figures.

1 answer:

5 0

<span>How many electrons would it take to equal the mass of a proton:
Here's one way of finding the value of it:
=> number of electrons is equivalent to 1 proton.
Let's have an example.
1.6726*10 -24g
_______________
1 proton
______________
9.109*10- ^28g
_______________
1 electron

Based on the given example above, the electrons is 1 839 per 1 proton.
It's about 1800 electrons/proton.</span>

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A leopard of mass 65kg climbs 7m up a tall tree. Calculate how much gravitational potential energy it gains. Assume g=10N/kg.

lisabon 2012 [21]

Answer: 4550 Joules

The formula for potential energy in gravitational field is as follows:

$E_{pot} = mgh=65kg \cdot 10 \frac{N}{kg} \cdot 7m = 4550 J$

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3 years ago

Vector A points along the + y axis and a magnitude of 100.0 units. Vector B points at an angle of 60.0 degrees above the +x axis

DedPeter [7]

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3 years ago

g An insulated piston–cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred

Andrews [41]

The question is not complete and the complete question is;

An insulated piston-cylinder device contains 4 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a P-Vdiagram with respect to saturation lines.

Answer:

Voltage of source = 175.33 V

I've attached an image of the P-Vdiagram of the process with respect to saturation lines.

Explanation:

To solve this question, we'll adopt the following assumptions.

- The kinetic and potential energy changes are negligible

-The thermal energy stored in the cylinder is negligible

-The cylinder is well insulated and thus heat transfer is negligible.

If the contents of the cylinder is taken as the system, the energy balance is;

Ein - Eout = ΔEsystem

Thus, [W(e-internal) + W(pw-internal)] - Wout = ΔU

Thus, IVΔt + W(pw-internal) = ΔH

= m(h2 - +1)

Now,looking at steam table A-5 attached to this answer, at P =175, x1 is calculated to be 0; dryness fraction, x2 = 0.5; vf = 0.001057m³/kg and hf = 487.01KJ/Kg and hg at saturation vapour = 2700.2

So, from this question v1=vf = 0.001057m³/kg and h1 = hf = 487.01KJ/Kg.

Also, h2 = hf + x2 (hg - hf) = 487.1 + 0.5(2700. 2 - 487.1) = 1593.65 KJ/Kg

The mass of the water is defined as;

M=V/v1

Volume(V) = 4L from the question. Since v1 is in m³/kg, let's convert V to m³. So V=4/1000 = 0.004

So, M = 0.004/0.001057 = 3.784kg

Now the formula for the voltage is ;

V = [M(h2 - h1) - W(pw-internal)] /(IΔt)

Δt=45minutes. We convert it to seconds to get 45x60 =2700 seconds

V = [[3.784 x (1593.65 - 487.1) - 400]/(8 x2700)] x 1000 = 175.33 V