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maria [59]
3 years ago
6

Zero field spot for opposite unequal charges

Physics
1 answer:
xz_007 [3.2K]3 years ago
7 0

Answer:

The zero field location has to be on the line running between the two point charges because that's the only place where the field vectors could point in exactly opposite directions. It can't be between the two opposite charges because there the field vectors from both charges point toward the negative charge.

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
2 years ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Stels [109]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

So

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now by integrating above equation

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
3 years ago
THIS IS A SCIENCE QUESTION IF THESE ARNT ANSWERED IM GOING TO FAIL SCIENCE
a_sh-v [17]

Answer:

1.a storm with a violent wind, in particular a tropical cyclone in the Caribbean.

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5.a rising of the sea as a result of atmospheric pressure changes and wind associated with a storm.

6.A hurricane dies down when it loses its energy source, which is usually warm water at the surface of the ocean.

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Explanation:

hope this helps

4 0
3 years ago
A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle
s344n2d4d5 [400]

Answer:

0.24 kgm²

Explanation:

L = length of the bat = 81.3 cm = 0.813 m

m  = mass of the bat = 0.96 kg

d  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

T  = Period of oscillation = 1.35 sec

I = moment of inertia of the bat

Period of oscillation is given as

T = 2\pi \sqrt{\frac{I}{mgd}}

1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}

I = 0.24 kgm²

6 0
3 years ago
A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface
Aloiza [94]

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, \sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa

Surface energy of magnesium oxide, SE = 1.0\ J/m^{2}

Modulus of elasticity of the material, E = 225 GPa = 225\times 10^{9}\ Pa

Now,

To calculate the maximum allowable surface crack length:

L = \frac{2E\times SE}{\sigma^{2}\pi }

L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm

7 0
3 years ago
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