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IRINA_888 [86]
3 years ago
13

What is the mass number of an atom with 3 protons, 4 neutrons, and 3 electrons

Physics
1 answer:
snow_tiger [21]3 years ago
5 0

Mass number = no. of protons + no. of neutrons

 so, it would be, 3+4 = 7

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An object weighs 10N on earth .what is the objects weight on a planet one tenth the earths mass and one half its radius?
earnstyle [38]
We know, weight = mass * gravity 
10 = m * 9.8
m = 10/9.8 = 1.02 Kg

Now, Let, the gravity of that planet = g'
g' = m/r²   [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)²   [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g   
g' = 2/5 * 9.8
g' = 3.92

Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N

In short, Your Answer would be 4 Newtons

Hope this helps!
3 0
3 years ago
A horizontal line above the time axis of a speed vs. time graph means an object is ___.
miv72 [106K]
A horizontal line on a speed/time graph means a constant speed.
6 0
3 years ago
Please help me with this question
snow_lady [41]

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6 0
3 years ago
Read 2 more answers
Differentiate between atmospheric pressure and pressure.​
Dovator [93]

Answer: atmospheric is air by the earth and pressure is just someone or something doing it

Explanation:

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

8 0
3 years ago
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