A 9 volt battery produces a current of 0.2A. What is the resistance?

Sending food to a country that has been hit by a hurricane is an example of which foreign policy tool?

sasho [114]

International Disaster Management or FEMA

8 0

3 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

2 years ago

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A car takes 15 minutes to travel along a road that is 20 km long.

iogann1982 [59]

Answer:

C

Explanation:

15. 20

x4. x4

60. 80

80km/h

8 0

2 years ago

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What is the formula for instant velocity when I have time and distance?

Tomtit [17]

Answer:

v(t)= (d/dt)x(t)

Explanation:

The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t. Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point t 0 is the rate of change of the position function, which is the slope of the position function

x ( t ) at t 0 .

3 0

2 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago