Which of the following would create voltage in a coil of wire?

Watt (w) is a drived unit why

ValentinkaMS [17]

Answer:

because it is from a mathematical combination of SI base units

Explanation:

4 0

3 years ago

Resistance of a light bulb with 0.33 A of current flowing from a 12V battery?

Sergio039 [100]

Resistance = (voltage) / (current)

Resistance = (12v) / (0.33 A)

Resistance = (12/0.33) ohms

<em>Resistance = 36.4 ohms</em>

8 0

3 years ago

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= $10+1.50$

= $11.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{11.5}$

$=26.08 \ s$

Throughout the opposite direction, when the boat seems to be travelling then,

= $10-1.50$

= $8.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v=\frac{v}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{8.5}$

$=35.29 \ s$

hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

8 0

3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

a

$v_2 = 5.6 \ m/s$

b

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

So

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

3 years ago

How do you find the net force acting on an object?

weqwewe [10]

Answer:

C. Add all the force vectors

Explanation:

The net force acting on an object is the vector sum of all the forces on the object.

Remember, Newton's first law tells us a body at rest will remain at rest or that in uniform motion will continue in motion unless acted by unbalanced forces.These unbalanced forces act in all direction towards the body thus to get the net force you require a summation of all these force with respect to their magnitudes and directions.

For example a force of 3N towards the East direction acting on a body and another force of 2N towards the West direction on the same body will generate a net force of 1N towards the East direction.

4 0

4 years ago