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2 years ago

Which two options describes behaviors of particles that are related to the chemical properties of the materials

Physics

1 answer:

jarptica [38.1K]2 years ago

3 0

Answer:

The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.

Explanation:

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Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed

makkiz [27]

Using the given formula with v0=56 ft/s and h=40 ft
h = -16t2 + v0t
40 = -16t2 + 56t
16t2 - 56t + 40 = 0
Solving the quadratic equation:
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32
We have two possible solutions
t1 = (56+24)/32 = 2.5
t2 = (56-24)/32 = 1
So initially the ball reach a height of 40 ft in 1 second.

3 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

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:

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

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

CHECK COMMENT FOR C

3 0

2 years ago

The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t^2 − 4.0t^3 m. Find (a) the velocity

KengaRu [80]

<h2>Answer:</h2>

(a) v(t) = [10.0t - 12.0t²] m/s and a(t) = [10.0 - 24.0t ] m/s² respectively

(b) -28.0m/s and -38.0m/s² respectively

(d) 0.83s

(e) x(t) = 1.1573 m [where t = 0.83s]

<h2>Explanation:</h2>

The position equation is given by;

x(t) = 5.0t² - 4.0t³ m --------------------(i)

(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;

v(t) = dx(t) / dt = $\frac{dx}{dt}$ = $\frac{d}{dt}$ [ 5.0t² - 4.0t³ ]

v(t) = 10.0t - 12.0t² --------------------------------(ii)

Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s

Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;

a(t) = dx(t) / dt = $\frac{dv}{dt}$ = $\frac{d}{dt}$ [ 10.0t - 12.0t² ]

a(t) = 10.0 - 24.0t --------------------------------(iii)

Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²

(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;

=> v(t) = 10.0t - 12.0t²

=> v(2.0) = 10.0(2) - 12.0(2)²

=> v(2.0) = 20.0 - 48.0

=> v(2.0) = -28.0m/s

Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;

=> a(t) = 10.0 - 24.0t

=> a(2.0) = 10.0 - 24.0(2)

=> a(2.0) = 10.0 - 48.0

=> a(2.0) = -38.0 m/s²

Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²

(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)

Therefore, substitute v = 0 into equation (ii) and solve for t as follows;

=> v(t) = 10.0t - 12.0t²

=> 0 = 10.0t - 12.0t²

=> 0 = ( 10.0 - 12.0t ) t

=> t = 0 or 10.0 - 12.0t = 0

=> t = 0 or 10.0 = 12.0t

=> t = 0 or t = 10.0 / 12.0

=> t = 0 or t = 0.83s

At t=0 or t = 0.83s, the position of the particle will be maximum.

To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;

<em>Substitute t = 0 into equation (i)</em>

x(t) = 5.0(0)² - 4.0(0)³ = 0

At t = 0; x = 0

<em>Substitute t = 0.83s into equation (i)</em>

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t) = 1.1573 m

At t = 0.83; x = 1.1573 m

Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s

(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s

(e) The maximum position function is found when t = 0.83s as shown in (c) above;

Substitute t = 0.83s into equation (i)

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t) = 1.1573 m [where t = 0.83s]

8 0

2 years ago

A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pres

marishachu [46]

Answer:

C) The pressure reading stays the same.

Explanation:

3 0

2 years ago

Read 2 more answers

If it takes an airplane 15 minutes to go from 30 mph to 330 mph, what is its acceleration?

zzz [600]

Using the a=vf-vi divided by tf-ti:
A is acceleration
Vf is final velocity- 330
Vi is intial velocity-30
Tf is final time-15
Ti is initial time-0
A = 330-30 divided by 15-0
A = 300 divided by 15
A= 20 m/s^2
Hope this helps

3 0

2 years ago