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3 years ago

Which part of the manufacturing process involves heating and rolling blocks of steel into flat sheets and bars?

Physics

1 answer:

faltersainse [42]3 years ago

8 0

A. Forming

Explanation:

Metal forming is a metallurgical process which involves heating and rolling blocks of steel into flat sheets and bars.

Metal forming is the process of working metals and utilizing its properties to form different shapes.

Metal forming is made possible by metallic bonds in metals.
The bulk of the physical properties of metals can be attributed to these bond types.
In metal forming, metals are shaped into different substances.
Metal forming preserves the mass of the metallic body and the shapes modified by force.

Learn more:

Metals brainly.com/question/9053565

#learnwithBrainly

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To make ice, a freezer that is a reverse Carnot engine extracts 37 kJ as heat at -17°C during each cycle, with coefficient of pe

Dvinal [7]

Answer:

a) 6.4 kJ

b) 43.4 kJ

Explanation:

$Q_{a}$ = Heat absorbed = 37 kJ

$β$ = Coefficient of performance = 5.8

$W$ = Work done

Heat absorbed is given as

$Q_{a}$ = $β$ $W$

37 = (5.8) $W$

$W$ = 6.4 kJ

$Q$ = work per cycle required

$Q$ = $Q_{a}$ + $W$

$Q$ = 37 + 6.4

$Q$ = 43.4 kJ

5 0

3 years ago

A single conservative force Fx = (6.0x  12) N (x is in m) acts on a particle moving along the x axis. The potential energy asso

jasenka [17]

Answer:

$U(3)=-43J$

Explanation:

Potential energy is minus the integral of Fdx. Doing the integration yields:

$U=\int\limits {6.0x+12}\, dx$

$U=-3x^2-12x+C$

$U(0)=20J$

$U(0)=-3(0)^2+12(0)+C=20$

$C=20$

Now for x=3.0m

$U(3)=-3*(3)^2-12(3)+20$

$U(3)=-43J$

6 0

3 years ago

A 795-loop square armature coil with a side of 10. 5 cm rotates at 70. 0 rev/s in a uniform magnetic field of strength 0. 45 t.

Agata [3.3K]

The rms voltage output of the generator is 1.94 × 10⁻ ⁵ V.

RMS is an acronym for root mean squared. An RMS value is more than just the "amount of AC power that causes the same heating impact as an analogous DC power" or something along those lines.

No. of loop = 795

Diameter of the coil = 10.5 cm

Radius of the coil = 5.25 cm

Magnetic Field, B = 0.45 T

Time, t = 70.0 rev/s

$V_{rms} =\frac{NwAB}{\sqrt{2} }$

Where,

N = No. of loop

A = Area of the coil

B = Magnetic Field

$V_{rms}$ = Voltage rms

Area of the coil = πr²

= 86.57 cm²

w = 2π/t

=( 2 × 3.141)/70.0

= 0.089

$V_{rms} =\frac{795*0.089*86.57* 0.45}{\sqrt{2} }\\\\V_{rms} =\frac{2756.36}{\sqrt{2} }\\\\\\V_{rms} =\frac{2756.36}{1.414 }\\\\V_{rms} = 1.94 * 10^-^5 V$

Therefore, the rms voltage output of the generator is 1.94 × 10⁻ ⁵ V.

Learn more about rms voltage here:

brainly.com/question/13156072

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8 0

8 months ago

What do all hydrogen atoms and ions have in common?

Leokris [45]

They have the same Number of protons

3 0

2 years ago

A spring-loaded gun can fire a projectile to a height h if it is fired straight up. if the same gun is pointed at an angle of 45

Naily [24]

When firing straight up:
v^2 = u^2 - 2gh, where v = final velocity = 0, u = initial velocity, g = gravitational acceleration, h = maximum height attained.

Then,
0 = u^2 - 2gh
u = Sqrt (2gh) ---- (1)

When firing at 45°,
Initial velocity, U = u Sin 45 = Sqrt (2gh)·Sin 45

Maximum height, H = U^2*(Sin Ф)^2/2g

substituting;
H = [Sqrt (2gh)·Sin 45]^2*(Sin 45)^2]/2g
H = [2gh*(Sin 45)^2*(Sin 45)^2]/2g
H = [h*(Sin 45)^4] = h/4

Therefore, maximum height when the gun fires at 45° is a quarter of maximum height when the gun fires vertically up.

6 0

3 years ago