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topjm [15]
3 years ago
9

(a) List the four classifications of steels. (b) For each, briefly describe the properties and typical applications.

Physics
1 answer:
tatiyna3 years ago
4 0

Answer:

Type of steels and their uses :

Type                            Carbon percentage                           uses

Low carbon steel         0.1% to 0.3 %           Food cans ,car body ,nut & bolt.

Medium carbon steel    0.3% to 0.70 %   Metal chains ,wheel rims,wire ropes

High carbon steel         0.7% to 1.3%          Hammers,drills ,dies and taps.

High speed steel         0.6 %                     Cutting tools in lathe machine.

Type                                           Properties

Low carbon steel     -          Fairly strong and easily get rusty              

Medium carbon steel -       Having high hardness than low carbon steel

High carbon steel  -            Having high hardness than medium carbon steel

High speed steel  -             Harder ,can retain its shape at high temperature

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the same with that of products

Explanation:

In a chemical reaction, the total charge of the reactants must be the same with that of products.

Charges must be conserved or balanced in chemical reactions.

  • In both acidic and basic/neutral medium electrons are used to balance the charge.
  • The appropriate number of electrons is added to the side with a larger charge.
  • One electron is used to balance each positive charge.
  • This ensures that the sum of charges on both sides the same.

Learn more:

Balanced equation brainly.com/question/5297242

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4 0
3 years ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
Which units are used to measure force?
shtirl [24]
Newtons are used to measure force
7 0
3 years ago
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A 10-cm-long wire is pulled along a u-shaped conducting rail in a perpendicular magnetic field. the total resistance of the wire
Debora [2.8K]

In the above case we can say that power given by external agent to pull the rod must be equal to the power dissipated in the form of heat due to magnetic induction.

Part a)

when we pull the rod with constant speed then power required will be product of force and velocity

here we will have

P = F.v

P = 4 W

v = 4 m/s

now we will have

4 = F*4

F = 1N

So external force required will be 1 N

PART B)

now in order to find magnetic field strength we can say

P = \frac{v^2B^2L^2}{R}

here we know that induced EMF in the wire is E = vBL

so power due to induced magnetic field is given by

P = \frac{E^2}{R}

4 = \frac{4^2*B^2*0.10^2}{0.20}

by solving above equation we will have

B = 2.24 T

5 0
3 years ago
A ball is thrown vertically downward from the top of a 30.6-m-tall building. The ball passes the top of a window that is 10.7 m
Gekata [30.6K]

Answer:

v = 19.6 m/s

Explanation:

Height of building = 30.6 m.

Height of window from the ground level= 10.7 m.

Acceleration due to gravity = 9.8 \frac{m}{s^2}

At initial condition ball at rest condition so u= 0 m/s.

Lets take when passes through the window ,velocity is v.

Here acceleration is constant so we can apply motion equation .

We know that

v= u + a t

So by putting the values

v = 0 +9.8 x 2

v = 19.6 m/s

So the velocity of ball is 19.6 m/s when passes through the window after 2 s.

7 0
3 years ago
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