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Tanzania [10]
4 years ago
6

A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density ρ is not un

iform but varies with the distance r from the center of the spherical charge distribution, according to the relationship ρ=βr for 0<=0<=a, where β is a positive constant, and ρ=0, and r>a.A. Show that the total charge Q in the spherical region of radius a is βπa^4B. In terms of β,r,a and fundamential constants, determine the magnitude of the electric field at a point a from distance r from the center of the spherical charge distribution for each of the following cases.i. r>aii. r=aiii 0
Physics
1 answer:
baherus [9]4 years ago
5 0

A negative charge and a positive charge

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B. Increases
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4 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

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4 years ago
What quantity is measured on the x-axis of a position vs. time graph.
tensa zangetsu [6.8K]

Answer:

Velocity

Explanation:

7 0
3 years ago
If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?
Arada [10]

The force on the fry is 0.015 N

Explanation:

We can find the force acting on the fry by using Newton's second law:

F=ma

where

F is the net force on the fry

m is its mass

a is its acceleration

For the fry in this problem,

m=3 g = 0.003 kg

a=5.0 m/s^2

Therefore, the force exerted on the fry is

F=(0.003)(5)=0.015 N

Learn more about Newton's second law here:

brainly.com/question/3820012

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3 years ago
HELP!! ALL MY POINTS WILL BE GIVEN
Marta_Voda [28]

Answer:

-6 m/s^2

Explanation:

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2 years ago
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