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2 years ago

If the element gallium has an atomic number of 31 and an atomic mass of 70, how many neutrons does it have?

Physics

1 answer:

Mumz [18]2 years ago

6 0

Answer:

The answer is 39.

Explanation:

The atomic number refers to the number of protons and the atomic mass is the sum of the protons and neutrons. So, you would just do 70 - 31 and that gets you 39.

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Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

3 years ago

A relatively small impact crater 20 kilometers in diameter could be made by a comet 2 kilometers in diameter traveling at 25 km/

nadya68 [22]

Answer:

$KE=1.3\times 10^{21}\ J$

$KE=3.1\times 10^{5}\ E$

Explanation:

(a)

Given:

mass of comet, $m=4.2\times 10^{12}\ kg$

velocity of the comet, $v=2.5\times 10^4\ m.s^{-1}$

Now, the kinetic energy of the comet can be given by:

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 4.2\times 10^{12}\times (2.5\times 10^4)^2$

$KE=1.3\times 10^{21}\ J$

(b)

Given:

energy released by 1 megaton of TNT, $E=4.2\times 10^{15}\ J$

Now the kinetic energy of the comet in terms of energy of 1 megaton TNT:

$KE=\frac{1.3\times 10^{21}}{4.2\times 10^{15}} E$

i.e.

$KE=3.1\times 10^{5}\ E$

4 0

3 years ago

The finished product or waste that forms as a result of a process is known as

Rasek [7]

The finished product or waste that forms as a result of a process is output.

6 0

3 years ago

Read 2 more answers

What are the three longest wavelengths for standing waves on a 264- cm -long string that is fixed at both ends

miv72 [106K]

The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:

5.2 meters.
2.6 meters.
1.7meters.

Given data:

Length of the fixed string = 264cms = 2.64 meters

The wavelength for standing waves is given by:

λ = 2L/n

where,

λ is the wavelength
L is the length of the string

For n = 1,

λ = 2×2.6/1

= 5.2 meters

For n = 2,

λ = 2×2.6/2

= 2.6 meters

For n = 3,

λ = 2×2.6/3

= 1.7 meters

To learn more about standing waves: brainly.com/question/14151246

#SPJ4

7 0

2 years ago

A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.

qaws [65]

Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444) = θ₂ - 22

77.70 = θ₂ - 22

θ₂ - 22 = 77.7

θ₂ = 77.7 + 22 = 99.7

So initial temperature before cooling ≈ 100°C . Option C.

5 0

3 years ago

Read 2 more answers