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Nookie1986 [14]
3 years ago
12

Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar

ound it. Its a new topic for me and am a bit confused how it works. The question seems simple but cant seem to get to the correct answer. The question:
An equation has three variables A,B and C.
A = B^2 + 2B^4/C . If the dimension of A is [L]^2/[T]^2, what must be the dimensions of B and C?

Options:
1. [B] = [L]^2/[T]^2 and [C] = [L]/[T]
2. [B] = [L]/[T] and [C] = [L]/[T]
3. [B] = [L]/[T] and [C] = (1/2)[T]/[L]
4. [B] = [L]/[T] and [C] = [T]/[L]
5. [B] = [L]/[T] and [C] = (1/2)[L]/[T]

Would be really helpful if someone could explain it. Thank you
Physics
1 answer:
omeli [17]3 years ago
7 0

Answer:

2. [B] = [L]/[T] and [C] = [L]/[T]

Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.

B² = [L]²/[T]²

B = [L]/[T]

B⁴/C² = [L]²/[T]²

C²/B⁴ = [T]²/[L]²

C² = B⁴ [T]²/[L]²

C² = ([L]/[T])⁴ [T]²/[L]²

C² = [L]²/[T]²

C = [L]/[T]

Notice we ignore the 2 coefficient, which is unitless.

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Identify two factors you must know to describe the motion of an object along a straight line
Sveta_85 [38]

In order to describe motion along a straight line, you must state the speed and direction of the motion. Those two quantities, together, comprise what's known as "velocity".

6 0
3 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
Help<br> How much current will flow when a 120 V power supply is connected to a 30<br> resistor ?
AVprozaik [17]
Current= voltage divided by resistance
120/30=4
7 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
4 years ago
a. When the electric field between the plates is 75% of the dielectric strength and energy density of the stored energy is 2800
Olegator [25]

Answer: The value of the dielectric constant k = 1.8

Explanation:

If C= ε A/d and

Electrostatic energy W = 1/2CV^2

Substitutes C in the first formula into the energy formula.

W = 1/2 ε A/d × V^2

Let us remember that electric field strength E is the ratio of potential V to the distance d. Where V = Ed

Substitute V = Ed into the energy W.

W = 1/2 × ε A/d ×( Ed )^2

W = 1/2 × ε A/d × E^2 × d^2

d will cancel one of the ds

W = 1/2 × ε Ad × E^2

W/Ad = 1/2 × ε × E^2

W/V = 1/2 × ε E^2

Where Ad = volume V

E = dielectric strength

εo = permittivity of free space = 8.84 x 10^-12 F/m

W/V = 2800 J/m^3

Let first calculate the dielectric strength

2800 = 1/2 × 8.84×10^-12 × E^2

5600 = 8.84×10^-12E^2

E^2 = 5600/8.84×10^-12

E = sqrt( 6.3 × 10^14)

E = 25 × 10^7

75% of E = 18.9 × 10^6Jm

The permittivity of the material will be achieved by using the same formula

2800 = 1/2 × ε E^2

2800 = 0.5 × ε × (18.9×10^6)^2

2800 = ε × 1.78 × 10^14

ε = 2800/1.78×10^14

ε = 1.57 × 10^-11

Dielectric constant k = relative permittivity

Relative permittivity is the ratio of the permittivity of the material to the permittivity of the vacuum in a free space. That is

k = 1.57×10^-11/8.84×10^-12

k = 1.776

k = 1.8 approximately

Therefore, the value of the dielectric constant k is 1.8

3 0
3 years ago
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