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3 years ago

13

1. the most suitable cutting process to use on ferrous metals such as straight carbon steel is

2 answers:

dangina [55]3 years ago

6 0

Answer:

plasma arc

cutting

Explanation:

It is plasma arc cutting because i think it is

swat323 years ago

4 0

Plasma so answer D.......

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What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

Alina [70]

Thank you for posting your question here at brainly. But your question seems incomplete. I will assume you based the situation below:

<span>An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T.

The </span> largest possible magnitude of the acceleration of the electron due to the magnetic field is <span>= 2.6 × 10 ¹⁶ m / s ²</span>

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3 years ago

The zinc plate is coated with mercury

raketka [301]

Amalgamating is the coating of zinc plate with mercury.

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3 years ago

The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler

guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

$v^{2}=u^{2}+ 2as$

$620^{2}=0^{2}+ 2\times a \times 0.89$

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.

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3 years ago

Select all of the statements that are true.

Strike441 [17]

I think its all four of them could be wrong but try all four !!!!!!

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3 years ago

The chain of length L and mass per unit length rho is released from rest on the smooth horizontal surface with a negligibly smal

devlian [24]

Answer:

Part a)

$a = \frac{x}{L} g$

Part b)

$T = \rho x g(1 - \frac{x}{L})$

Part c)

$v = \sqrt{gL}$

Explanation:

Part a)

Net pulling force on the chain is due to weight of the part of the chain which is over hanging

So we know that mass of overhanging part of chain is given as

$m = \rho x$

now net pulling force on the chain is given as

$F = \rho x g$

now acceleration is given as

$F = Ma$

$\rho x g = \rho L a$

$a = \frac{x}{L} g$

Part b)

Tension force in the part of the chain is given as

$mg - T = ma$

$\rho x g - T = \rho x a$

$\rho x(g - a) = T$

$\rho x (g - \frac{x}{L} g) = T$

$T = \rho x g(1 - \frac{x}{L})$

Part c)

velocity of the last link of the chain is given as

$a = \frac{x}{L} g$

$v\frac{dv}{dx} = \frac{x}{L} g$

now integrate both sides

$\int v dv = \frac{g}{L} \int x dx$

$\frac{v^2}{2} = \frac{gL}{2}$

$v = \sqrt{gL}$

3 0

2 years ago