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ss7ja [257]
3 years ago
6

What is the differential equation governing the growth of current in the circuit as a function of time after t=0? express the ri

ght-hand side of the differential equation for di(t)dt in terms of i(t), vb, r, and l?
Physics
1 answer:
Reika [66]3 years ago
3 0

Answer:

v_{b}=ir+L\frac{di}{dt}

Explanation:

A differential equation that contain a term with di(t)/dt is in a RL circuit. Here we have

v_{b}=v_{r}+v_{i}

where vr is the voltage in the resistance, vi is the voltage in the inductance and vb is the source voltage. But also we have that

v_{r}=ir\\v_{i}=L\frac{di}{dt}

where L is the inductance of the circuit, r is the resistance an i is the current. By replacing we have the differential equation

v_{b}=ir+L\frac{di}{dt}

I hope this is useful for you

regards

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An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

(a)   Formula to calculate the density is as follows.

            \rho = \frac{Q}{\frac{4}{3}\pi a^{3}}

                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

                     = 2.42 \times 10^{-2} C/m^{3}

Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

7 0
3 years ago
Please help asap! Thank you.
olganol [36]

Answer:

direct current

Explanation:

it has a direct path to go down to reach the specific point

4 0
2 years ago
Read 2 more answers
1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th
Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f  + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

  final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4  ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a)  final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5  ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a)  final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀  ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a)  final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

3 0
3 years ago
Dr. Aloysius found four homogeneous mixtures, but he thinks that only three are alloys. He tests each one and determines their e
Lina20 [59]

Answer:

A. carbon and boron

Explanation:

Carbon and boron is not an alloy.

An allow forms between metals and metals using their huge electron could.

Carbon is a non-metal, boron is a also a non-metal

Two non-metals combining together does not make an alloy.

Iron, nickel, aluminum  are all metals.

6 0
2 years ago
Read 2 more answers
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0
3 years ago
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