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Julli [10]
3 years ago
5

How many neutrons are in an atom of carbon with a mass number of 13

Physics
1 answer:
BabaBlast [244]3 years ago
4 0

C is the answer to your question.

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When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi
Nikitich [7]

The resistance of the cylindrical wire is R=\frac{\rho l}{A}.

Here R is the resistance, l is the length of the wire and A is the area of cross section. Since the wire is cylindrical A=\frac{\pi d^2}{4} .

Comparing two wires,

R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}

Dividing the above 2 equations,

\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{A_2 }{A_1}  \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2}  \frac{d_2^2 }{d_1^2}  \\

Since d_2=2d_1

The above ratio is

\frac{R_1}{R_2}=\frac{1.68(10^{-8})  }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264

We also have,

\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0
3 years ago
A heat engine takes in 840 kJ per cycle from a heat reservoir. Which is not a possible value of the engine's heat output per cyc
zhannawk [14.2K]
We have by the first law of thermodynamics tha energy is preserved, hence we cannot have over 840kJ per cycle. We have by the laws of thermodynamics (the 2nd one in specific) that the entropy of a system cannot increase. We cannot have an output of  840 kJ per cycle from a heat engine because then that would mean that the entropy would stay the same, while any heat engine increases it. Hence, any value \geq 840 kJ is acceptable.
8 0
3 years ago
Which three characteristics do mechanical waves and electrocmagnetic
almond37 [142]
B the waves transfer energy from its source
8 0
2 years ago
What is the acceleration due to gravity at an altitude of 116?
Schach [20]
Gravitational acceleration (Ga) is  inversely proportional to  k / Distance^2

so Ga * Distance^2 = K

On the surface of Earth acceleration due to gravity is about 9.8m/s^2 with an average distance to the earths core of about 6371 km (Wolfram alpha).

So k = 9.8 * 6371^2
I'm presuming that your distance of 116 is km

As 
Ga = k / distance^2 

Ga = ((9.8 * 6371^2) / (6371 + 116)^2 ) = 397778481.8 / 42081169

= 9.45 m/s^2 to 2sf
7 0
3 years ago
Sherlock Holmes is using an 9.99-cm-focal-length lens as his magnifying glass. To obtain maximum magnification, where must the o
Karolina [17]

Answer:

The object must be placed just ahead of focus towards the lens.

Explanation:

As we know that a magnifying glass is a convex lens and for the convex lens and for a convex lens we get the image virtual, erect and magnified only when it is placed between the optical center and the focus of the lens.

In that range when the object is nearest to the focus of the lens it has the maximum size.

So it can be possible if the object is placed at 9.98 cm from the optical center.

The largest image is formed at infinity when the object is placed at focus but it is not visible to the normal eyes.

5 0
3 years ago
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