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Serhud [2]
3 years ago
14

Select all that apply. Which of the following would increase the conductivity of a connecting wire?

Physics
1 answer:
lord [1]3 years ago
4 0
What are your choices 

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3. a) If the distance between two objects increases by a factor of 5 (5),
irina [24]

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

3 0
2 years ago
Explain why average velocity in one dimension can be positive or negative.
Softa [21]
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since velocity =  \frac{change in position}{time}, and time is always positive, velocity will be positive in one direction and negative in the other direction.
3 0
3 years ago
Read 2 more answers
a current of 250 amps is flowing through a copper wire with resistance of 2.09 x 10^4 ohms. what is the voltage
Dima020 [189]

V = I · R

Voltage = (current) · (Resistance)

Voltage = (250 A) · (2.09 x 10⁴)

Voltage = 5,225,000 volts .

I may be out of line here, but I'm pretty sure
that the resistance is 2.09 x 10⁻⁴ .
Then

Voltage = 0.05225 volt (not 5 million and something)
7 0
3 years ago
The mass of the Moon is 7.35 x 1022 kg, while that of Earth is 5.98 x 1024 kg. The average distance from the center of the Moon
Kryger [21]

Answer:

aaa

Explanation:

m_e = Mass of the Earth =  5.98 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r_1 = Distance from the center of the Moon to the center of Earth = 6371000 m

r_2 = Distance from the center of the earth center to sun center

m_m = Mass of moon = 7.35\times 10^{22}\ kg

M = Mass of sun = 1.989\times 10^{30}\ kg

F_1=G\frac{m_em_m}{r_1^2}\\\Rightarrow F_1=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 7.35\times 10^{22}}{(384000000)^2}\\\Rightarrow F_1=1.988\times 10^{20}\ N

F_2=G\frac{Mm_e}{r_1^2}\\\Rightarrow F_2=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 1.989\times 10^{30}}{(149.6\times 10^9+6371000+695.51\times 10^6)^2}\\\Rightarrow F_2=3.511\times 10^{22} N

\frac{F_1}{F_2}=\frac{1.988\times 10^{20}}{3.511\times 10^{22}}\\\Rightarrow \frac{F_1}{F_2}=0.00566\\\Rightarrow F_1=F_20.00566

Hence the force of moon on earth is 0.00566 times the force of earth on moon center to center

4 0
3 years ago
Find the minimum value of n in the Balmer series for which the predicted wavelength is in the ultraviolet region of the spectrum
d1i1m1o1n [39]

Answer:

 λ =365.4 nm

Explanation:

Boh's atomic model of the Hydrogen atom the energy of each level is

        Eₙ = - 13.606 / n²

where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.

An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.

                 ED = 13.606 (1 / n₀² - 1 /n_{f}^{2})

we are going to apply this relationship to answer slash.

 

At the beginning of the studies of atomic transitions, each group did not consider having a different name

name        Initial state

Lymman         1

Balmer           2

the final state is any other state sta the continuum that corresponds to n = inf

Let's look for the highest energy of the Balmer series

              ΔE = 13.606 (1/2² - 1 /∞)

              ΔE = 3.4015 eV

Let's use the Planck relation for the energy

                E = h f = h c /λ

                λ = h c / E

Let's reduce the energy to J

              E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹

            λ = 6.63 10⁻³⁴  3 10⁸ / 5.4424 10⁻¹⁹

            λ = 3.654 10⁻⁷ m

            λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)

            λ =365.4 nm

this eta radiation in the ultraviolet range

5 0
3 years ago
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