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ss7ja [257]
3 years ago
15

The moment of inertia of a uniform equilateral triangle with mass m and side length a about an axis through one of its sides and

parallel to that side is (1/8) ma^2. What is the moment of inertia of a uniform regular hexagon of mass m and side length a about an axis through two opposite vertices?
Physics
1 answer:
Bond [772]3 years ago
8 0
No Idea sir skskskhhhhfffvftujbvf
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Which of the following correctly describes the relationship between current and voltage as the voltage of a battery increases. R
Y_Kistochka [10]

Answer:_COC1\/2+_H\/2O>_HC1+CO\/2

Explanation:

Need help asap

3 0
3 years ago
The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the
wel

Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C

5 0
3 years ago
Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge
asambeis [7]

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

E =\dfrac{kq}{r^2}

q =\dfrac{Er^2}{k}

inserting all the values

q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}

      q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0
3 years ago
what is the wavelength λ2λ2lambda 2 of the second laser that would place its second maximum at the same location as the fourth m
nordsb [41]

The distance separating two wave crests is known as the wavelength. The wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Double slit interference (constructive) can be described by the equation dsinθ=m λ

Here, the order is (m), the wavelength is (), and the slit distance is (d).

Information provided-

The laser's wavelength is d/8.

The slit separation is 0.500 mm in length.

Put the distance into consideration to determine the wavelength values as,

λ1=0.5 x 10⁻³/8

λ1=0.0625 x 10⁻³

Due to the second laser's wavelength, its second maximum would be at the same spot as the first laser's fourth minimum.

Therefore, we must first determine the fourth order minima as,

dsinθ= (4-1/2) λ1

dsinθ= (7/2) λ1

Using the second maxima, dsinθ= 2λ₂

Put the following value for in the equation above:

7/2λ1 = 2λ₂

7/2 x (0.0625 x 10⁻³) = 2λ₂

λ₂=0.000109375m.

Therefore, the wavelength of the second laser, where its second maximum would be located, is 0.109375 mm.

Learn more about wavelength here-

brainly.com/question/13533093

#SPJ4

4 0
1 year ago
If 20 beats are produced within a single second, which of the following frequencies could possibly be held by two sound waves tr
zhuklara [117]

The correct choice is

D. 22 Hz and 42 Hz.

In fact, the beat frequency is given by the difference between the frequencies of the two waves:

f_B = |f_1 -f_2|

In this problem, the beat frequency is f_B=20 Hz, therefore the only pair of frequencies that gives a difference equal to 20 Hz is

D. 22 Hz and 42 Hz.

4 0
3 years ago
Read 2 more answers
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