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3 years ago

9.00 V is applied to a wire with a resistance of 52.0 ohm. At what distance from the wire is the magnetic field 2.22x 10^-8 T?

Physics

1 answer:

Romashka-Z-Leto [24]3 years ago

6 0

Answer:

r = 1.56 m

Explanation:

B = μ₀I/2πr

r = μ₀I/2πB

I = 9.00 / 52.0 = 0.173 A

μ₀ = 4πe-7 T•m/A

r = 4πe-7(0.173)/2π(2.22e-8)

r = 2e-7(0.173)/(2.22e-8)

r = 1.5592... ≈ 1.56 m

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Shalnov [3]

Answer:

No.

Explanation:

Given the following :

Velocity (V) of ball = 5m/s

Radius = 1m

Can the ball reach the highest point of the circular track

of radius 1.0 m?

The highest point in the track could be considered as the diameter of the circle :

Radius = diameter / 2;

Diameter = (2 * Radius) = (2*1) = 2

Maximum height which the ball can reach :

Using the relation :

Kinetic Energy = Potential Energy

0.5mv^2 = mgh

0.5v^2 = gh

0.5(5^2) = 9.8h

0.5 * 25 = 9.8h

12.5 = 9.8h

h = 12.5 / 9.8

h = 1.2755

h = 1.26m

Therefore maximum height which can be reached is 1.26m.

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7 0

3 years ago

Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

4 0

3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

3 years ago

What is the mechanical advantage of the screw shown below? O A. 14.1 O B. 2 O C. 12.6 O D. 8.2.

Vilka [71]

Answer: C. 12.6

Explanation: 2*pi*1.8= 11.304

11.304/0.9= 12.56

3 0

2 years ago

Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in t

noname [10]

Answer:

A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.

Explanation:

In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.

Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.

The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.

The phenomenon of Doppler effects is generally applicable to both sound and light.

An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.

Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.

3 0

3 years ago