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miskamm [114]
3 years ago
13

Government as an academic field of study ​

Physics
2 answers:
Ira Lisetskai [31]3 years ago
4 0

Answer: True

Explanation:

Government as an academic field of study refers to the study of political institutions, processes and theories taught in schools, colleges and higher institution of learning. It is called Political Science in higher institution.

Savatey [412]3 years ago
3 0
Government as an academic field of study is a vast subject, encompassing political institutions and political theories. ... Government allows for civil order by creating laws and protecting its citizens. In order to ensure enforcement of laws, government also creates punishments for those who break the law.
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A piece of copper wire with thin insulation, 200 m long and 1.00 mm in diameter, is wound onto a plastic tube to form a long sol
bearhunter [10]

Answer:

 N= 3

Explanation:

For this exercise we must use Faraday's law

          E = - dФ / dt

         Ф = B . A = B Acos θ

tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives

         E = - A cos θ (B - B₀) / t

The angle enters the magnetic field and the normal to the area is zero

         cos 0 = 1

         A = π r²

   

In the length of the wire there are N turns each with a length L₀ = 2π r

          L = N (2π r)

          r = L / 2π N

    we substitute

          A = L² / (4π N²)

The magnetic field produced by a solenoid is

           B = μ₀ N/L   I

for which

            B₀ = μ₀  N/L   I

           

The final field is zero, because the current is zero

            B = 0

We substitute

           E = - (L² / 4π N²)  (0 - μ₀ N/L I) / t

           E = μ₀ L I / (4π N t)

           N = μ₀ L I / (4π t E)

The electromotive force is E = 0.80 mV = 0.8 10⁻³ V

let's calculate

           N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]

           N  = 320 10⁻⁷ / 9.6 10⁻⁶

           N = 33.3 10⁻¹

          N= 3

           

7 0
2 years ago
मारवतन गनुहास (What Is MKS Syste<br>Convert 5 solar days into second.)​
mr_godi [17]

Answer:

5 Days to Seconds = 432000

Explanation:

7 0
3 years ago
A cars engine can deliver 300,000 watts of power to its wheels.
harkovskaia [24]

Answer:

A. 1,800,000 J

B. 4473.87 N

C. 3.728 m/s²

5 0
2 years ago
A 10-cm-long wire is pulled along a u-shaped conducting rail in a perpendicular magnetic field. the total resistance of the wire
Debora [2.8K]

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.

Part a)

when we pull the rod with constant speed then power required will be product of force and velocity

here we will have

P = F.v

P = 4 W

v = 4 m/s

now we will have

4 = F*4

F = 1N

So external force required will be 1 N

PART B)

now in order to find magnetic field strength we can say

P = \frac{v^2B^2L^2}{R}

here we know that induced EMF in the wire is E = vBL

so power due to induced magnetic field is given by

P = \frac{E^2}{R}

4 = \frac{4^2*B^2*0.10^2}{0.20}

by solving above equation we will have

B = 2.24 T

5 0
3 years ago
Suppose that you make a series RC circuit with a capacitor and a known resistor that has a 5% tolerance: R= 5.20 ± 0.26kΩ. You p
Masja [62]

Answer:

correct answer is C

Explanation:

The time constant of an RC circuit is

           τ = RC

so to find the capacitance

          C = τ/ R

          C = 2.150 / 5.20 10³

          C = 4.13 10⁻⁴ F

to find the error we use the worst case

         ΔC = | |\frac{dC}{d \tau }| \ \Delta  \tau + | \frac{dC}{dR} | \ \Delta R

the absolute value guarantees that we find the worst case, we evaluate the derivatives

          ΔC = 1 /R Δτ + τ/R²  ΔR

the absolute values ​​of the errors are

          Δτ = 0.002 s

          ΔR = 0.3 kΩ

we substitute

           ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)²   0.3 10³

           ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵

           ΔC = 1.77 10⁻⁵ F

the uncertainty or error must be expressed with a significant figure

            ΔC = 2 10⁻⁵ F

the percentage error is

            Er% =\frac{\Delta C}{C} \ 100

            Er% = \frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100

            Er% = 4.8%

the correct answer is C

3 0
3 years ago
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