All categories

3 years ago

A coil of wire can become a temporary magnet if connected to a battery O True O False

Physics

1 answer:

GrogVix [38]3 years ago

4 0

Answer:

true

Explanation:

An Electromagnet is called as a temporary magnet, because electromagnet is made up of insulated copper wire wrapped around soft iron, which allows current to flow in it and at that moment iron gains it magnetic properties and when the current is switched off, then there will be no flow of current and hence iron loses its magnetic properties.

Hence it works temporarily as long as electric current flows in a conductor and hence known as temporary magnet. .

You might be interested in

A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp

Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m, m2 = 4m, v1=vf_p, v2 = vf_alpha

The conservation momentum states that:

pi = pf

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

m1 = m, m2 = 4m, q1 = e, q2 = 2e

and v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 = k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0

3 years ago

Which of the following is not considered a major work flow structure?

mixas84 [53]

Answer:

The correct option is;

D. Fabrication

Explanation:

A workflow flow is a detailed business process consisting of a series of required interconnected tasks in directed graph format that is executable by workflow management system.

Considering each of the options, we have

A. Work center

This consists of part of the transformation input to output. The location

B. Project

This is the unique identifier of the task to be processed

C. Assembly line

Forms part of the required input where transformation takes place and items are being processed within the assembly line

D. Fabrication

Here the item is fixed, without motion, therefore this is not considered a major work flow structure

E. Continuous flow

Here again, the items are being processed and are in motion, which constitutes a workflow structure.

8 0

2 years ago

4. A 1,000 kg truck moving at 10 m/s runs into a concrete wall. It takes 0.5 seconds for the truck to conipietery

Maru [420]

Answer:

$\large \boxed{\text{h. 20 000 N}}$

Explanation:

Force is the change in momentum over time

F = Δp/Δt

1. Calculate the change in momentum

p₁ = mv₁ = 1000 kg × 10 m/s = 10 000 kg·m·s⁻¹

p₂ = 0

Δp = p₂ - p₁= (0 - 10 000) kg·m·s⁻¹ = -10 000 kg·m·s⁻¹

2. Calculate the force

$\begin{array}{rcl}F & = & \dfrac{\Delta p}{\Delta t}\\\\& = & \dfrac{-10 000 \text{ kg$\cdot$m$\cdot$ s}^{-1}}{\text{ 0.5 s}}\\\\& = & \textbf{-20 000 N}\\\end{array}\\\text{The negative sign shows that the force is exerted opposite to the direction of motion.}\\\text{The magnitude of the force is $\large \boxed{\textbf{20 000 N}}$}$

3 0

3 years ago

Read 2 more answers

If both mass and velocity of a ball are tripled, the kinetic energy is increased by a factor of:.

hichkok12 [17]

Answer:

the answer is 27

Explanation:

1/2mv^2 = KE

if mass is 3, and velocity is 3 (3^2 = 9)

then KE increases by a factor of 27

5 0

2 years ago

When the current in a toroidal solenoid is changing at a rate of 0.0275 a/s, the magnitude of the induced emf is 12.4 mv. when t

slavikrds [6]

Answer:

41 turns

Explanation:

$\dfrac{d\phi}{dt}$ = Induced emf = 12.4 mV

$\dfrac{dI}{dt}$ = Current changing rate = 0.0275 A/s

L= Inductance

$\phi$ = Average flux = 0.00285 wb

N = Number of turns

Change in flux is given by

$\dfrac{d\phi}{dt}=L\dfrac{dI}{dt}\\\Rightarrow 12.4\times 10^{-3}=L0.0275\\\Rightarrow L=\dfrac{2.4\times 10^{-3}}{0.0275}\\\Rightarrow L=0.08727\ H$

Flux through each turn is given by

$\dfrac{\phi}{N}=L\dfrac{I}{N}\\\Rightarrow 0.00285=0.08727\times \dfrac{1.34}{N}\\\Rightarrow N=\dfrac{0.08727\times 1.34}{0.00285}\\\Rightarrow N=41.03221\ turns$

The number of turns is 41

3 0

3 years ago