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Dimas [21]
3 years ago
10

Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having

the dimension of time. Determine the dimen- sions of the constants A and B. (b) Determine the dimen- sions of the derivative dx/dt 5 3At2 1 B.
Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) \frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
3 years ago
Read 2 more answers
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field
Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

E_{mid} = E_{right} +E_{left}\\\\E_{right}  = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt}  = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }  - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0
3 years ago
Compare and contrast the theories about the origin of the universe
Vaselesa [24]

Answer:

The best-supported theory of our universe's origin centers on an event known as the big bang. This theory was born of the observation that other galaxies are moving away from our own at great speed in all directions, as if they had all been propelled by an ancient explosive force.

Explanation:

hope this helps tho i don't quite know what you mean

7 0
3 years ago
You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
5 0
4 years ago
Only 35 % of the intensity of a polarized light wave passes through a polarizing filter. What is the angle between the electric
Nana76 [90]

Answer:

The angle between the electric field and the axis of the filter is 54⁰

Explanation:

Apply the equation for intensity of light through a polarizer.

I = I_oCos^2 \theta

where;

I is the intensity of the transmitted light

I₀ is the intensity of the incident light

θ is the incident angle

If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;

\frac{I}{I_o}  = Cos^2 \theta\\\\\frac{35}{100} =  Cos^2 \theta\\\\Cos^2 \theta = 0.35\\\\Cos\theta = \sqrt{0.35} \\\\Cos\theta = 0.5916\\\\\theta = Cos^{-1}(0.5916)\\\\\theta  = 54 ^0

Therefore, the angle between the electric field and the axis of the filter is 54⁰

3 0
3 years ago
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