Ask question

Ask question

All categories

2 years ago

10

A microscope illuminator uses a transformer to step down the 120 V AC of the wall outlet to power a 12.0 V,50 W microscope bulb.

What is the resistance of the bulb filament

1 answer:

Anna35 [415]2 years ago

5 0

Answer:2.88 ohms

Explanation:

R= V^2 / P

12^2/50

144/50

2.88 ohms

You might be interested in

6-What is the difference between the critical point and the triple point?

hichkok12 [17]

Answer:

The triple point represents the combination of pressure and temperature that facilitates all phases of matter at equilibrium. The critical point terminates the liquid/gas phase line .

Explanation:

8 0

3 years ago

Read 2 more answers

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

3 years ago

The part of a circuit that carries the flow of electrons is referred to as the?

Oksanka [162]

Answer:

Conductor

Explanation:

Current is carried by a conductor.

__

The purpose of a dielectric and/or insulator is to prevent current flow. An electrostatic field may set up the conditions for current flow, but it carries no current itself.

7 0

3 years ago

You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.

allochka39001 [22]

Answer:

Explanation:

I hope this helps!

3 0

3 years ago

A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the

sergeinik [125]

Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = $\frac{weight}{area}$

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = $\frac{5000}{4.45}$

= 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

7 0

3 years ago