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4 years ago

What has research determined about the orbit of an electron around a nucleus?

1 answer:

4 0

<span>I think the correct answer from the choices listed above is the second option. The research that determined about the orbit of an electron around a nucleus would be that electrons orbit the nucleus in circular motions exactly as shown on the classic Bohr model.</span>

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A 27.0-m steel wire and a 48.0-m copper wire are attached end to end and stretched to a tension of 145 N. Both wires have a radi

algol13

Answer:

The time taken by the wave to travel along the combination of two wires is 458 ms.

Explanation:

Given that,

Length of steel wire= 27.0 m

Length of copper wire = 48.0 m

Tension = 145 N

Radius of both wires = 0.450 mm

Density of steel wire $\rho_{s}= 7.86\times10^{3}\ kg/m^{3}$

Density of copper wire $\rho_{c}=8.92\times10^{3}\ kg/m^3$

We need to calculate the linear density of steel wire

Using formula of linear density

$\mu_{s}=\rho_{s}A$

$\mu_{s}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{s}=7.86\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{s}=5.00\times10^{-3}\ kg/m$

We need to calculate the linear density of copper wire

Using formula of linear density

$\mu_{c}=\rho_{s}A$

$\mu_{c}=\rho_{s}\times\pi r^2$

Put the value into the formula

$\mu_{c}=8.92\times10^{3}\times\pi\times(0.450\times10^{-3})^2$

$\mu_{c}=5.67\times10^{-3}\ kg/m$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{s}=\sqrt{\dfrac{T}{\mu_{s}}}$

$v_{s}=\sqrt{\dfrac{145}{5.00\times10^{-3}}}$

$v_{s}=170.3\ m/s$

We need to calculate the velocity of the wave along the steel wire

Using formula of velocity

$v_{c}=\sqrt{\dfrac{T}{\mu_{c}}}$

$v_{c}=\sqrt{\dfrac{145}{5.67\times10^{-3}}}$

$v_{c}=159.9\ m/s$

We need to calculate the time taken by the wave to travel along the combination of two wires

$t=t_{s}+t_{c}$

$t=\dfrac{l_{s}}{v_{s}}+\dfrac{l_{c}}{v_{c}}$

Put the value into the formula

$t=\dfrac{27.0}{170.3}+\dfrac{48.0}{159.9}$

$t=0.458\ sec$

$t=458\ ms$

Hence, The time taken by the wave to travel along the combination of two wires is 458 ms.

4 0

4 years ago

A light bulb provides a resistance of 20 Ω to the 30 A current that runs through it. Determine the voltage of the battery in the

likoan [24]

Explanation:

Ohm's law:

V = IR

V = (30 A) (20 Ω)

V = 600 V

4 0

3 years ago

Read 2 more answers

A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?

fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as $V_1$ , the relation with the final volume as

$V_2 = V_1 +0.163\% V_1$

$V_2 = V_1 +0.00163V_1$

$V_2 = 1.00163V_1$

Initial temperature = $21.1\°C$

Let T be the temperature after expanding by the formula of volume expansion

we have,

$V_2 = V_1 (1+\gamma \Delta t)$

Where $\gamma$ is the volume coefficient of copper $5.1*10^{-5}/C$

$1.00163V_1 = V_1(1+\gamma(T-21.1\°))$

$1.00163 = 1+5.1*10^{-5}(T-21.1\°)$

$0.00163 = 0.000051T-0.0010761$

$T = 53.0608\°C$

Therefore the temperature is 53.06°C

7 0

4 years ago

A coating is being applied to reduce the reflectivity of a pane of glass to light with a wavelength of 522 nm incident near the

fredd [130]

Answer:

t = 94.91 nm

Explanation:

given,

wavelength of the light = 522 nm

refractive index of the material = 1.375

we know the equation

c = ν λ

where ν is the frequency of the wave

c is the speed of light

$\nu= \dfrac{c}{\nu\lambda}$

$\nu = \dfrac{3\times 10^8}{522 \times 10^{-9}}$

ν = 5.75 x 10¹⁴ Hz

the thickness of the coating will be calculated using

$t = \dfrac{\lambda}{4\mu_{material}}$

$t = \dfrac{522 \times 10^{-9}}{4\times 1.375}$

t = 94.91 nm

the thickness of the coating will be equal to t = 94.91 nm

7 0

3 years ago

A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?

taurus [48]

Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

$V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]$

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

8 0

3 years ago