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SpyIntel [72]
3 years ago
15

A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the paper. Directed out of the plane of the paper is a magn

etic field of 0.06 T. The field out of the paper decreases to 0.02 T in 12 milliseconds. What is the direction of the current induced?
Physics
1 answer:
Maslowich3 years ago
5 0

Explanation:

It is given that,

Number of turns, N = 200

Area of cross section, A = 8.5 cm²

Magnetic field is directed out of the paper and is, B = 0.06 T

The magnetic field is  out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Since, \epsilon=IR

I is the induced current

I=-\dfrac{N}{R}\dfrac{d\phi}{dt}

According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.

Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

8 0
3 years ago
An airplane with an airspeed of 120 km/h has a heading of 30 degree east of North in a wind that is blowing toward the east at 6
lesya692 [45]

Answer:

Explanation:

Velocity of plane relative to ground V_pg = ?

Given the velocity in vector form ,

velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j

V_wg = 60 i

V_pg = V_pw +V_wg

= 120 cos30 i + 120sin30j + 60i

= 164 i + 60 j

magnitude

=251 km / h

=

8 0
3 years ago
a certain car travels 20 km east then turns south for 13 km finally the car turns east again for 6 km
aliya0001 [1]

a^2+b^2=c^2

20km^2+13km^2=c^2

400km^2+169km^2=c^2

23.853720...km

7 0
3 years ago
Read 2 more answers
The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and
Salsk061 [2.6K]
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

Let:

b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°

We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
 
5 0
3 years ago
Weekend A<br> Assignment<br> Differentiate between forced and damped oscillation
4vir4ik [10]

Answer:

A damped oscillation means an oscillation that fades away with time while Forced oscillations occur when an oscillating system is driven by a periodic force that is external to the oscillating system.

Explanation:

Damping is the reduction in amplitude (energy loss from the system) due to overcomings of external forces like friction or air resistance and other resistive forces. ... When a body oscillates by being influenced by an external periodic force, it is called forced oscillation.

<h2><em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>helped</u></em><em><u> </u></em></h2>

<em><u>Welcome</u></em><em><u> </u></em>

6 0
3 years ago
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