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4 years ago

6

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

ion for E=kQ/r2 for r>R and E=kQr/R3 for r
Only need C & D. It asks for specfic points that I don't know how to find. I just know the general shape of the graph.

1 answer:

AlladinOne [14]4 years ago

8 0

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

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Answer

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$\omega = \dfrac{2\pi}{T}$

$\omega = \dfrac{2\pi}{ 86400 }$

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a) at equator

$v = R_E \omega$

$v = 6.38 \times 10^6\times 7.272 \times 10^{-5}$

v = 464 m/s

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$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)$

$a = 0.03374 m/s^2$

b) at a latitude of 61.0 ° north of the equator.

$R = R_E cos \theta$

$R = 6.38 \times 10^6\times cos 61^0$

$R = 3.093 \times 10^6 m$

$v = R \omega$

$v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}$

$v = 225 m/s$

acceleration of the person

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3 years ago

In the diagram below, if A's magnitude is 16 N and B's is 25 N, what is the magnitude of C?

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Assuming that’s a right triangle, in this case A^2 + C^2 = B^2 … (16)^2 + C^2 = (25)^2 … C = 19.2 N

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2 years ago

The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the

Ivanshal [37]

Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

(c) $\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) $a = 22.458\ m/s^{2}$

Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

$omega'$ = final angular velocity

$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

$\alpha = - 1.32\ rev/m^{2}$

Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

$\theta = 13674\ rev$

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

$\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) The radial acceleration is given by:

$\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s$

Linear acceleration is given by:

$a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}$

$a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}$

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3 years ago

A rock of mass 0.340 kg is spun horizontally at the end of a wire that has a diameter of 1.00 mm. If the wire gets stretched by

borishaifa [10]

Answer:

Y = 78.13 x 10⁹ Pa = 78.13 GPa

Explanation:

First we will find the centripetal force acting on the wire as follows:

F = mv²/r

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F = Force = ?

m = mass of rock = 0.34 kg

v = speed = 19 m/s

r = length of wire

Therefore,

F = (0.34)(19)²/r

F = 122.74/r

now, we find cross-sectional area of wire:

A = πd²/4

where,

A = Area = ?

d = diameter of wire = 1 mm = 0.001 m

Therefore,

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A = 7.85 x 10⁻⁷ m²

Now, we calculate the stress on wire:

Stress = F/A

Stress = (122.74/r)/(7.85 x 10⁻⁷)

Stress = 1.56 x 10⁸/r

Now, we calculate strain:

Strain = Δr/r

where,

Δr = stretch in length = 2 mm = 0.002 m

Therefore,

Strain = 0.002/r

now, for Young's modulus (Y):

Y = Stress/Strain

Y = (1.56 x 10⁸/r)/(0.002/r)

<u>Y = 78.13 x 10⁹ Pa = 78.13 GPa</u>

8 0

3 years ago