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3 years ago

Which two statements explain reasons to use a computer program to encode

Physics

2 answers:

Paraphin [41]3 years ago

8 0

Answer:a

Explanation:why not

mixas84 [53]3 years ago

3 0

Answer: A and B

Explanation: Because it makes the data you have secure and can only be authorized by you.

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A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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A man has a mass of 85kg, what is his weight

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The answer is he weighs 187.39 LBS/Pounds

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Which statements describe the motion of the object represented by the graph? Check all that apply.

Sav [38]

Answer:a, d, e

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2 years ago

(1) In nondestructive testing, a discontinuity may be defined as an interruption in the normal physical structure or configurati

QveST [7]

Answer: BOTH ARE TRUE

Explanation: Nondestructive testing or Evaluation is a term used in the field of science and technology to describe the evaluations, analysis or testing carried out on components of materials without destroying any part or components of the test materials. It is very useful in scientific research or industrial engineering environments. When any disruption of physical structure or configuration of a component will lead to discontinuing of the test, and it may not affect the usefulness of the affected parts.

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