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vova2212 [387]
2 years ago
13

James yap run fast while dribbling the ball as he crosses the middle part of the basketball court he changes his phase to avoid

opponent what specific
component of physical fitness did James Yap?​
Physics
1 answer:
DerKrebs [107]2 years ago
3 0

Answer:

The answer is "Speed".

Explanation:

In the question, the Speed is the correct answer because it can be viewed as its rate from, which a length is covered via an object. It is also divided by time into other components of distance. Its SI unit of speed seems to be the meter/second, but still, the kilometer/hour or, in the Us UK, miles per hour is the most basic fundamental of speed in everyday use.

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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
Find the angle (above the horizontal) at which a projectile achieves its maximum range, if y=y0.
KatRina [158]
The answer is 45 degrees. 
According to the Kinematics of projectile motion, if the purpose is to maximize range, optimum angle of landing is always 45 degrees.If the purpose is to maximize range & projection height is zero, the optimum angle of projection (and landing) is 45 degrees.
5 0
2 years ago
Ever tried to stop a 150-pound (68 kg) cannonball fired towards you at 30 mph (48 km/hr.)? No, probably not. But you may have tr
alex41 [277]
The two situations are similar because in both you are trying to minimize the damage and make the best out of a bad situation
8 0
2 years ago
If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 12m/s, how far from the base of the dorm should
Hitman42 [59]

Answer:

3.6 m

Explanation:

let x = horizontal distance between emily and allison should be for allison to catch the ball

Find horizontal speed of the ball

vx = 12 sin 30 = 12 x 0.5 = 6 m/s

To find time taken, we will use vertical values of the ball motion

Initial velocity in vertical direction

u = 12 cos 30 = 10.392 m/s

let a = g = 9.8m/s2

Use equation of motion

s = ut +1/2at^2

s = vertical distance = 8

8 = (10.392)t + (1/2)(9.8)t^2

8 = (10.392)t + (4.9)t^2

4.9t^2 + 10.392t - 8 = 0

Using formula of quadratic or calculator, we'll find

t = 0.6 and t = -2.72

We pick t=0.6s since it's not logical time in negative

Assuming no air resistance or external forces, the ball will move 6m/s horizontally. Hence using the formula of speed

speed vx = distance x / time

x = (vx)(t)

  = 6 x 0.6

  = 3.6 m

6 0
2 years ago
Buffalo, New York, experienced a snowstorm November 13–21, 2014. Residents refer to the event as “Snowvember.” What was the like
scZoUnD [109]
I know you're probably done with this by now, but the answer is *Lake-Effect Snow*
5 0
2 years ago
Read 2 more answers
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