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4 years ago

Why are moving iron instrument are of repulsion type

Physics

1 answer:

never [62]4 years ago

5 0

Answer:

he spring provides the controlling torque. The air friction induces the damping torque, which opposes the movement of the coil. The repulsion type instrument is a non-polarized instrument, i.e., free from the direction of current passes through it. Thus, it is used for both AC and DC

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A 2 kg toy cart and a 6 kg toy cart have a spring compressed between them. When the spring expands, it sends the 2 kg toy cart o

harkovskaia [24]

Answer:

The speed of second toy cart is 4 m/s.

Explanation:

Given that,

Mass of first toy cart = 2 kg

Mass of second toy cart = 6 kg

Speed of first toy cart = 12 m/s

We need to calculate the speed of second toy cart

Using formula of momentum

$m_{1}v_{1}=m_{2}v_{2}$

Where, m₁ = mass of first toy cart

m₂ = mass of second toy cart

v₁ = velocity of first toy cart

v₂ = velocity of second toy cart

Put the value into th formula

$2\times12=6\times v_{2}$

$v_{2}=\dfrac{2\times12}{6}$

$v_{2}=4\ m/s$

Hence, The speed of second toy cart is 4 m/s.

(c) is correct option

4 0

4 years ago

Đoạn dây dẫn thẳng có dòng điện I chạy qua, đặt trong từ trường

umka21 [38]

A straight piece of wire with a current I flowing through it is placed in a magnetic field

A straight piece of wire with a current I flowing through it is placed in a magnetic fielduniform and perpendicular to the magnetic field lines. Magnetic force acting on the string

A straight piece of wire with a current I flowing through it is placed in a magnetic fielduniform and perpendicular to the magnetic field lines. Magnetic force acting on the stringthere is a way

4 0

3 years ago

If an automobile's velocity changes from 25 m/s to 15m/s in 2s, then what is its acceleration?

tekilochka [14]

Hey!

u=25m/s
v=15m/s
t=2s
a=?

v-u=at
15-25=a×2
-10=2a
-10/2=a
a=-5m/s^2

Hope it helps...!!!

8 0

4 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

4 years ago

Why does a bouncing ball rise to a lower height with each bounce? What energy conversion is taking place?

dsp73

So we want to know why the does a bouncing ball rise to a lower height with each bounce. So lets say the ball is first on some height h. There it has potential energy Ep=m*g*h. Then as the ball starts falling to the ground the energy converts to kinetic energy Ek=(1/2)*m*v^2. When the ball falls to the ground, the kinetic energy transforms to elastic energy because the ball deforms as it hits the ground and some small quantity of heat. The heat goes to the air and to the ground so it gets removed from the system. So there is less energy in the system to be converted back to kinetic energy as the ball starts to rise in height again. Thats why the ball is not able to get bact to the same height as it started from.

7 0

4 years ago