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3 years ago

What is eletro magnet

Physics

1 answer:

Leona [35]3 years ago

7 0

A soft metal core made into a magnet by the passage of electric current through a coil surrounding it.

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Below is a physics question

Allisa [31]

Explanation:

i1=1ampere

i1+i2=I

VA−VB=i1×3Ω=i2×6

⇒1×3=i2×6

2i=21ampere

Hence I=i1+i2=1+0.5=1.5ampere

Req=2+3+63×6=2+93×6=4Ω

equivalent circuit

Refer image .2

From KVL

1req=v

v=1.5×4

8 0

3 years ago

Before scientists come up with a scientific question, what do they have to do first?

Mademuasel [1]

Scientist would check what other scientists have said before they come up with their own scientific question. That way they can record the results from the others and compare them with their own findings when they conduct their own experiments and record the findings.

8 0

3 years ago

Consider a human population that consumes a certain resource. Which of the following statements about this situation is most lik

Papessa [141]

If the size of the population increases, the amount of the resource consumed will also increase

3 0

3 years ago

Read 2 more answers

The following position vs time graph shows the motion of an object along its path from -10 to seconds. Based on this graph, what

Marrrta [24]

∆x (displacement) = v - u

where, v is the final position and u is the initial position.

Given, the final position of the object is 0 m and the initial position is 5 m.

∆x (displacement) = v - u

= 0 m - 5 m

= -5 m

Therefore, <u>C: -5m</u> is the correct answer.

6 0

3 years ago

A bullet of mass 0.016 kg traveling horizontally at a speed of 280 m/s embeds itself in a block of mass 3 kg that is sitting at

ivanzaharov [21]

(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

$p_i = p_f\\m u_b + M u_B = (m+M)v$

where

m = 0.016 kg is the mass of the bullet

$u_b = 280 m/s$ is the initial velocity of the bullet

M = 3 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

$v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s$

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J$

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

$K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J$

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).

4 0

4 years ago