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Mandarinka [93]
4 years ago
15

The amount of charge flowing through a particular point in a conductor is represented by the equation Q = at^3 + bt + c, where a

= 3.00 A/s^2, b = 5.00 A, and c = 8.00 A*s. Determine the current density at t = 0.700 s if the cross-sectional area of the conductor at the point of consideration is 2.50 cm^2. A/m^2

Physics
1 answer:
stira [4]4 years ago
3 0

Explanation:

Below is an attachment containing the solution.

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Where is visible light located on the electromagnetic spectrum and why?
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(Hypothetical question)<br> If the earth was square and not round what would it be like
Nuetrik [128]

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It would be a square and 2d

Explanation:

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2 years ago
Read 2 more answers
What is the speed of the object between 60 to 70 seconds?
lyudmila [28]

We have that the speed of the object between 60 to 70 seconds

V=\frac{d}{65}

From the question we are told

The speed of the object between 60 to 70 seconds

Generally the equation for Average speed   is mathematically given as

V=\frac{d}{t}

Where

t=\frac{60+70}{2}\\\\t=65seconds

Therefore

V=\frac{d}{65}

In conclusion

The speed of the object between 60 to 70 seconds

V=\frac{d}{65}

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

5 0
3 years ago
A comet is approaching Venus on a parabolic path with perigee distance of 18,200 km . Calculate the total time of travel (in hou
Anna35 [415]

Answer:

<em>The time traveled is 1.39 hrs</em>

Explanation:

Equation of Trajectory of a comet is given as

r=\frac{h^2}{\mu}\frac{1}{1+cos \theta}

Here

  • h is the specific angular momentum given as

                                           h=v_p r_p

  • μ is gravitational parameter whose value is 3.24859 \times 10^{14} \, \, m^3/s^2 for Venus
  • r_p is the perigee distance of parabolic which is 18200 km
  • As the path of comet is parabolic so energy is conserved i.e

                                 \frac{1}{2}m_c v_p^2-\frac{\mu m_c}{r_p}=0\\v_p=\sqrt{\frac{2 \mu}{r_p}}

So h is given as

                                    h=v_pr_p\\\\h=\sqrt{2 \mu r_p}\\h=\sqrt{2 \times 3.24859 \times 10^{14} \times 18200 \times 10^3}\\h=1.087 \times 10^{11}

So for point a where r=24500 km

                                r_1=\frac{h^2}{\mu}\frac{1}{1+cos \theta_1}\\24500 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_1}\\0.6731=\frac{1}{1+cos \theta_1}\\1+cos \theta_1=\frac{1}{0.6731}\\cos \theta_1=1.4857-1\\cos \theta_1=0.4857\\\theta_1=cos^{-1}0.4857\\\theta_1=1.0636 rad

So for point a where r=39000 km

                              r_2=\frac{h^2}{\mu}\frac{1}{1+cos \theta_2}\\39000 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_2}\\1.0715=\frac{1}{1+cos \theta_2}\\1+cos \theta_2=\frac{1}{1.0715}\\cos \theta_2=0.9332-1\\cos \theta_2=-0.0667\\\theta_2=cos^{-1}(-0.0667)\\\theta_2=1.6375 rad

So as per the Barkers equation

                      t_1-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_1+\frac{D_1^3}{3})

where

                      D_1=tan (\theta_1/2)\\D_1=tan(0.5318)\\D_1=0.5883

                   t_2-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3})

where

                    D_2=tan (\theta_2/2)\\D_1=tan(0.8187)\\D_2=1.0690

So

t_2-t_1=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3}-D_1+\frac{D_1^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}((1.0690)+\frac{(1.0690)^3}{3}-(0.5883)-\frac{(0.5883)^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}(0.8201)\\t_2-t_1=(6092)(0.8201)\\t_2-t_1=4996.21 s\\t_2-t_1=1.39 hrs\\

So the time traveled is 1.39 hrs

3 0
3 years ago
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