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malfutka [58]
2 years ago
7

First find ∮RB⃗ ⋅dl⃗ , the line integral of B⃗ around a loop of radius R located just outside the left capacitor plate. This can

be found from the usual current due to moving charge in Ampère's law, that is, without the displacement current. Find an expression for this integral involving the current I(t) and any needed constants given in the introduction.
Physics
1 answer:
Allisa [31]2 years ago
5 0

Answer:

the expression of current in the loop enclosed to the left of the capacitor plate is

I(t) = \frac{1}{\mu_0}\int B. dL

Explanation:

As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path

So we will have

\int B. dL = \mu_0 I(t)

so we have

I(t) = \frac{1}{\mu_0}\int B. dL

so above is the expression of current in the loop enclosed to the left of the capacitor plate

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If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
2 years ago
A block is projected with speed v across a horizontal surface and slides to a stop due to friction. The same block is then proje
natka813 [3]

Answer: C. The case on the inclined surface had the least decrease intotal mechanical energy.

Explanation:

First and foremost, it should be noted that the mechanical energy is the addition of the potential and the kinetic energy.

From the information given, it should be known that when the block is projected with the same speed v up an incline where is slides to a stop due to friction, the box will lose its kinetic energy but there'll be na increase in the potential energy as a result of the veritcal height. This then brings about an increase in the mechanical energy.

Therefore, the total mechanical energy of the block will decrease the least when the case on the inclined surface had the least decrease intotal mechanical energy.

8 0
1 year ago
Consider two waves defined by the wave functions y1(x,t)=0.50msin(2π3.00mx+2π4.00st) and y2(x,t)=0.50msin(2π6.00mx−2π4.00st). Wh
guapka [62]

Answer:

They two waves has the same amplitude and frequency but different wavelengths.

Explanation: comparing the wave equation above with the general wave equation

y(x,t) = Asin(2Πft + 2Πx/¶)

Let ¶ be the wavelength

A is the amplitude

f is the frequency

t is the time

They two waves has the same amplitude and frequency but different wavelengths.

4 0
2 years ago
Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by
mrs_skeptik [129]
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R  as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick T_R=T whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that T_L=-T

8 0
2 years ago
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
2 years ago
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