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rodikova [14]
3 years ago
15

An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch

arge density σ = -0.35 μC/m2. A thin wire, with linear charge density λ = 1.1 μC/m, is inserted along the shells' axis. The shell and the wire do not touch and these is no charge exchanged between them.a) What is the new surface charge density, in microcoulombs per square meter, on the inner surface of the cylindrical shell?b) What is the new surface charge density, in microcoulombs per square meter, on the outer surface of the cylindrical shell?c) Enter an expression for the magnitude of the electric field ou

Physics
1 answer:
gayaneshka [121]3 years ago
8 0

Answer:

Explanation:

Solution is in the picture attached

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For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod
Scorpion4ik [409]

Answer: -\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

2Br^.\rightarrow Br_2

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate=-\frac{d[Br^.]}{2dt}

or Rate=+\frac{d[Br_2]}{dt}

Thus -\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}

4 0
3 years ago
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Amiraneli [1.4K]

Answer:

a)      K = 2/3 π G m ρ R₁³ / R₂ ,  b) U = - G m M / r

Explanation:

The law of universal gravitation is

     F = G m M / r²

Part A

Let's use Newton's second law

     F = m a

The acceleration is centripetal

     a = v² / R₂

     

      G m M / R₂² = m v² / R₂

      v² = G M / R₂

They give us the density of the planet

    ρ = M / V

    V = 4/3 π R₁³

    M =   ρ V

    M =   ρ 4/3 π R₁³

    v² = 4/3 π G  ρ R₁³ / R₂

    K = ½ m v²

    K = ½ m (4/3 π G ρ R₁³ / R₂)

    K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

     F = - dU / dr

     ∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

    U- U₀ = G m M ∫ dr / r²

    U - U₀ = G m M (- r⁻¹)

We evaluate for

    U - U₀ = -G m M (1 / r_{f} -  1 /r_{i})

They indicate that for ri = ∞     U₀ = 0

    U = - G m M / r

6 0
3 years ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo
liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}

hence, the mean velocity is 8.4 m/s

8 0
3 years ago
What is the term to describe the rate of flow of electricity?
SVEN [57.7K]
The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.
7 0
3 years ago
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