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2 years ago

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A volcanic island arc is formed when _____. A. two oceanic plates converge and form a subduction zone B. a continental plate mee

ts an oceanic plate and form a subduction zone C. volcanic activity occurs within a plate D. three convergent plates form an intraplate gap

1 answer:

MakcuM [25]2 years ago

6 0

Answer:

C. volcanic activity occurs within a plate.

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A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it

natka813 [3]

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c} }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

8 0

2 years ago

If the 80 ohm resistor fails, will the 50 ohm and 100 ohm resistor continue to operate? Why or why not?

sleet_krkn [62]

Answer:

No, the 50 ohm and 100 ohm resistor will not continue to operate.

Explanation:

A closed circuit is the circuit in which there is no break between the negative and the positive end of the battery.

When in this, combinational circuit the 80 ohm resistor fail then there will not any continue supply of current in the circuit due to the breakage because the electron will flow from negative end of the battery to positive end if their is no breaking in the circuit.

Therefore the 50 ohm and 100 ohm circuit will not continue to operate because of the breaking of the circuit and current will not flow.

7 0

2 years ago

A tank contains 100 gal of water and 50 oz of salt.water containing a salt concentration of 1 4 (1 1 2 sin t) oz/gal flows into

Alchen [17]

Answer:

Explanation:

Heres the possible full question and solution:

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito the tank at a rate of 2 gal/min, and the mixture in the tank flows out at the same rate.

a. Find the amount of salt in the tank at any time.

b. Plot the solution for a time period long enough so that you see the ultimate behavior of the graph.

c. The long-time behavior of the solution is an oscillation about a certain constant level. What is this level? What is the amplitude of the oscillation?

solution

a)

Consider the tank contains 100gal of water and 50 oz of salt

Assume that the amount of salt in the tank at time t is Q(t).

Then, the rate of change of salt in the tank is given by $\frac{dQ}{dt}$ .

Here, $\frac{dQ}{dt}$ =rate of liquid flowing in the tank - rate of liquid flowing out.

Therefore,

$Rate_{in} =2gal/min \times \frac{1}{4} (1+ \frac{1}{2}sin t)oz/gal\\\\\\ \frac{1}{2} (1+ \frac{1}{2}sin t)oz/min\\\\\\Rate_{out}=2gal/min \times\frac{Q}{100}oz/gal\\\\\frac{Q}{50}oz/min$

Therefore,

$\frac{dQ}{dt}$ can be evaluated as shown below:

$\frac{dQ}{dt}=\frac{1}{2}(1+\frac{1}{2}\sin t)-\frac{Q}{50}\\\\\\\frac{dQ}{dt}+\frac{1}{50}Q=\frac{1}{2}+\frac{1}{4}\sin t$

The above differential equation is in standard form:

$\frac{dy}{dt}+Py=G$

Here, $P=\frac{1}{50},G=\frac{1}{2}+\frac{1}{4}\sin t$

The integrating factor is as follows:

$\mu(t)=e^{\int {P}dt}\\\mu(t)=e^{\int {\frac{1}{50}}dt}\\\mu(t)=e^{\frac{t}{50}}$

Thus, the integrating factor is $\mu(t)=e^{\frac{t}{50}}$

Therefore, the general solution is as follows:

$y\mu(t)=\int {\mu (t)G}dt\\\\Qe^{\frac{t}{50}}=\int {e^{\frac{t}{50}}(\frac{1}{2}+\frac{1}{4}\sin t) dt}\\\\Qe^{\frac{t}{50}}=\frac{1}{2}\int {e^{\frac{t}{50}}dt + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}\\\\\Qe^{\frac{t}{50}}=25 {e^{\frac{t}{50}} + \frac{1}{4}\int {\sin t {e^{\frac{t}{50}}} dt}+C...(1)$

Here, C is arbitrary constant of integration.

Solve $\int {\sin te^{\frac{t}{50}}} dt}$

Here $u = e^{\frac{t}{50}}$ and $v =\sin t$ .

Substitute u , v in the below formula:

$\int{u,v}dt=u\int{v}dt-\int\frac{du}{dt}\int{v}dt\dot dt\\\\\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{1}{50}\int{e^{\frac{t}{50}}\cos t}dt...(2)$

Now, take $u = e^{\frac{t}{50}}$ , $v =\sin t$

Therefore, $\int{e^{\frac{t}{50}}\cos t} dt=\int {e^{\frac{t}{50}}\sin t}dt - \frac{1}{50}\int{e^{\frac{t}{50}}\sin t}dt...(3)$

Use (3) in equation(2)

$\int {e^{\frac{t}{50}}\sin t}dt=-e^{\frac{t}{50}}\cos t + \frac{e^{\frac{t}{50}}}{50}\sin t - \frac{1}{2500}\int{e^{\frac{t}{50}}\sin t}dt\\\\\frac{2501}{2500}\int{e^{\frac{t}{50}}\sin t}dt={e^{\frac{t}{50}}\cos t}+\frac{e^{\frac{t}{50}}}{50}\sin t\\\\\int{e^{\frac{t}{50}}\sin t}dt=\frac{2500}{2501}{e^{\frac{t}{50}}\cos t}+\frac{50}{2501}e^{\frac{t}{50}}\sin t...(4)$

Use (4) in equation(l) .

$Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+C$

Apply the initial conditions $t =0, Q = 50$ .

$50=25-\frac{625}{2501}+c\\\\c=\frac{63150}{2501}$

So, $Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}$

Therefore, the amount of salt in the tank at any time is as follows:

$Qe^{\frac{t}{50}}=25 e^{\frac{t}{50}} - \frac{625}{2501}e^{\frac{t}{50}}\cos t +\frac{25}{5002}e^{\frac{t}{50}}\sin t+\frac{63150}{2501}e^{\frac{-t}{50}}$

b)

sketch the solution curve as shown in attachment as graph 1:

CHECK COMMENT FOR C

3 0

2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

Tema [17]

<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

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2 years ago