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3 years ago

Please help

Physics

2 answers:

sveticcg [70]3 years ago

8 0

Answer:

A and E

Explanation:

Time and cost..........

Fiesta28 [93]3 years ago

3 0

Answer:A and E

Explanation:just did it

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A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

4 years ago

What is 1.3 for PE=mgh and for KE=1(m)v^2 and E=PE+KE and V

emmainna [20.7K]

Answer:

1.3 is equal to m times v to the second times PK, meaning the answer to your question is 3.3 K

Explanation:

Have a gday!

8 0

3 years ago

Consider two ideal gases, A and B, at the same temperature. The rms speed of the molecules of gas A is twice that ofgas B. How d

katen-ka-za [31]

The rms speed of the molecules of gas A is twice that of gas B. The molecular mass of A is one fourth to that of B.

Answer: Option B

<u>Explanation:</u>

Measuring the speed of particles at a given point in time results in a large distribution of values. Some molecules can move very slowly, others very fast, and because they are still moving in different directions, the speeds may be zero. (Velocity, vector quantity that corresponds to the speed and direction of the molecule.)

To correctly estimate the average velocity, you must take the squares of the mean velocity and take the square root of this value. This is known as the root mean square (rms) velocity and is shown as follows:

$V_{r m s}=\sqrt{\frac{3 R T}{M}}$