Answer it or not your choice

An elastic conducting material is stretched into a circular loop of 11.2 cm radius. It is placed with its plane perpendicular to

anygoal [31]

Answer:

0.426 volts

Explanation:

It is given that,

The radius of a circular loop, r = 11.2 cm = 0.112 m

An elastic conducting material is stretched into a circular loop.

It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s

We need to find the emf induced in the loop at that instant.

$\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{d}{dt}(BA)\\\\=\dfrac{d}{dt}(\pi r^2 B)\\\\=\pi B\dfrac{d}{dt}(r^2)\\\\=2\pi B r\dfrac{dr}{dt}\\\\=2\pi \times 0.88\times 0.112\times 0.688\\\\=0.426\ V$

So, the magnitude of induced emf is 0.426 volts.

4 0

2 years ago

what is the position of centre of curvature for concave and convex mirror show with the help of diagram if you can

Anon25 [30]

it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram

8 0

2 years ago

STATION 1

jok3333 [9.3K]

Chedda chease snadwhik

8 0

2 years ago

A 60 kg man jumps down from a 0.8 m table. What is the speed when he

Rom4ik [11]

Answer: Speed = 4 m/s

Explanation:

The parameters given are

Mass M = 60 kg

Height h = 0.8 m

Acceleration due to gravity g= 10 m/s2

Before the man jumps, he will be experiencing potential energy at the top of the table.

P.E = mgh

Substitute all the parameters into the formula

P.E = 60 × 9.8 × 0.8

P.E = 470.4 J

As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.

When hitting the ground,

K.E = P.E

Where K.E = 1/2mv^2

Substitute m and 470.4 into the formula

470.4 = 1/2 × 60 × V^2

V^2 = 470.4/30

V^2 = 15.68

V = square root (15.68)

V = 3.959 m/s

Therefore, the speed of the man when hitting the ground is approximately 4 m/s

4 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

2 years ago