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3 years ago

How do scientific tests help determine the properties of substances

Physics

1 answer:

tino4ka555 [31]3 years ago

7 0

It can help determine substances that appear similar but react differently under the same circumstances.

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Transmission Lines and Health. Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the elec

Blababa [14]

Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:

B = μ₀I/(2πr)

B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line

Given values:

μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m

Plug in and solve for B:

B = 4π×10⁻⁷(170)/(2π(8.0))

B = 4.25×10⁻⁶T

The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:

4.25×10⁻⁶/(5.0×10⁻⁵)

= 0.085

= 8.5%

The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.

8 0

4 years ago

A researcher claiming that females were more empathetic than males would test that hypothesis by using __________ statistics.

Alex17521 [72]

A researcher claiming that females were more empathetic than males would test that hypothesis by using inferential statistics.

6 0

4 years ago

Read 2 more answers

A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. De

Len [333]

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

$\int\limits^a_b {x} \, dx$

a=2

b=1

x=5÷(3s^(1/3)+s^(5/2)) m/s^2

dx=dv

Integrate the left side the standard method.

$\int\limits^a_b {x} \, dx$

a=v

b=0

dx=dv

Integrating

=v^2/2

Use Simpson's rule for the right site.

$\int\limits^a_b {x} \, dx$

a=b

b=a

x=f(x)

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

=0.8376

Solve for v.

v=1.295

5 0

3 years ago

How does the electric force between two charged particles change if one particle's charge is increased by a factor of 2?

Burka [1]

if one of the charge is doubled, the electric potential energy would be doubled too

8 0

3 years ago

Read 2 more answers

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an elec

kicyunya [14]

Answer:

B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

Em₀ = U = q V

final point. Right out of the electric field

em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

q V = ½ m v²

v = $\sqrt{2qV/m}$

we calculate

v = $\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }$

v = $\sqrt{632.3353 \ 10^8}$

v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

F = m a

the force is

F = q v x B

as the velocity is perpendicular to the magnetic field

F = q v B

acceleration is centripetal

a = v² / r

we substitute

qvB =1/2 m v² / r

B = v $\frac{m v}{2 q r}$

we calculate

B = $\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}$

B = 1.1413 10⁻² T

8 0

3 years ago