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valentina_108 [34]

2 years ago

15

A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

ded in order for the magnetic force to balance the weight of the rod?

1 answer:

slamgirl [31]2 years ago

8 0

Answer:

$i = 4.9 A$

Explanation:

Force on a current carrying rod due to magnetic field is given as

$F = iLB$

here we know that

$i =$ current in the rod

$B = 0.10 T$

$L = 1.0 m$

now magnetic force is balanced by the weight of the rod

so we will have

$iLB = mg$

$i(1.0)(0.10) = 0.05 \times 9.8$

$i = 4.9 A$

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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

A car travels 30 miles in 1 hour on a winding mountain road. Which of the following is a true statement?

siniylev [52]

Answer:

The true statement is:

"(C) The magnitude of the average velocity is equal to 30 m.p.h."

Explanation:

Given that a car travels 30 miles in 1 hour on a winding mountain road.

Let' check all the statements one by one:

(A) The magnitude of the total displacement is larger than the distance traveled.

Since the entire motion of the car is not exactly given in the question, so it is not possible to tell whether the magnitude of the total displacement is larger than the distance traveled or not.

Thus, this statement is not true.

(B) The magnitude of the average velocity is greater than 30 m.p.h.

The average velocity of an object is defined as the total displacement covered by the particle divided by the total time taken in covering that displacement.

Total distance covered by the car = 30 miles.

Total time taken by the car to cover this distance = 1 hour.

Therefore, the average velocity of the car for this time interval = $\rm \dfrac{30\ miles}{1\ hour }= 30\ m.p.h.$

Thus, this statement is also not true.

(C) The magnitude of the average velocity is equal to 30 m.p.h.

As is cleared in part (B) section above, the average velocity of the car in the given time interval is 30 m.p.h.

Thus, this statement is true.

(D)The magnitude of the average velocity is less than to 30 m.p.h.

Since. the average velocity of the car is 30 m.p.h.

Thus, this statement is not true.

(E)The car traveled with a constant speed of 30 m.p.h.

The motion of the car on the mountain road is not thoroughly given in the question, so again it is not possible to tell whether the car traveled with a constant speed of 30 m.p.h. or not.

Thus, this statement is also not true.

4 0

2 years ago

Read 2 more answers

you are sitting on a beach and a wave strikes the shore every 10 seconds. a surfer tells you that these waves travel at a speed

aev [14]

Answer:

50 meters

Explanation:

5 0

2 years ago

The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴ × 1000 m

r = 1.08 × 10⁷ m, G = 6.67 × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08 × 10⁷ )²/75( 6.67 × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0

2 years ago

Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (

snow_lady [41]

Answer

given,

net charge = +2.00 μC

we know,

1 coulomb charge = 6.28 x 10¹⁸electrons

1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron

= 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

= 1.256 x 10¹³ electrons

since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

6 0

2 years ago