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weeeeeb [17]
3 years ago
15

An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete

rmine the electric field everywhere.
Physics
1 answer:
xenn [34]3 years ago
3 0

Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.

As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

for a<r<b er have

E*2π*r*L= Q inside/εo       in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L

E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

E=ρ*(r-a)/εo

Finally for r>b

E*2π*r*L =ρ*2*π*b* (b-a)*L/εo

E=ρ*b* (b-a)*/r*εo

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A 68 kg deer has a momentum of 952 kg∙m/s. What is its velocity?
Gwar [14]

All we need to do is divide the momentum and weight to find the velocity:

v = m / w

Solve using the values we are given.

v = 952/68 = 14 m/s

Best of Luck!

5 0
3 years ago
Problem:
kenny6666 [7]

Answer:

a) x = 1.5 *10⁻⁴cos(524πt) m

b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s

   a =  -1.5 *10⁻⁴(524π)²cos(524πt) m/s²

c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

Explanation:

x = Acos(ωt)

ω = 2πf = 2π(262) = 524π rad/s

x = 1.5 *10⁻⁴cos(524πt)

v = y' = -Aωsin(ωt)

v = -1.5 *10⁻⁴(524π)sin(524πt)

a = v' = -Aω²cos(ωt)

a =  -1.5 *10⁻⁴(524π)²cos(524πt)

not sure about the last part as time is generally not given in mm

I will show at 1 second and at 0.001 s to try to cover bases

x(1) = 1.5 *10⁻⁴cos(524π(1))

x(1) = 1.5 *10⁻⁴cos(524π)

x(1) = 1.5 *10⁻⁴(1)

x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm

x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))

x(0.001) = 1.5 *10⁻⁴cos(0.524π)

x(0.001) = 1.5 *10⁻⁴(-0.0753268)

x(0.001) = -1.129902...*10⁻⁵ m

x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm

7 0
3 years ago
3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed rel
kramer

Answer:

v_s=\frac{v}{1+\frac{m}{M} }

Explanation:

Let the shells speed with respect to ground be v_s

Let the shells speed with respect to ground be v_g

mass of shell, m

mass of gun, M

The relative velocity of shell with respect to a free unconstrained gun, v

According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.

So, the relation between the relative velocity and their individual velocity will be:

v=v_s+v_g ......................(1)

<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

M.v_g-m.v_s=0

substitute the value of v_g from equation (1)

M\times (v-v_s)=m\times v_s

M.v-M.v_s=m.v_s

M.v=M.v_s+m.v_s

v_s=\frac{M.v}{(M+m)}

v_s=\frac{v}{(\frac{(M+m)}{M}) }

v_s=\frac{v}{1+\frac{m}{M} }

5 0
3 years ago
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
OlgaM077 [116]

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F  mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

8 0
3 years ago
A light bulb will glow when electrons flow through it. As the electron flow increases, the brightness increases as well. A stude
Inga [223]
The correct answer is: <span>Unscrew one light, if the others remain on it is a parallel circuit.</span>
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