So the transformer in this problem really doesn't matter. If the lamp is using energy at the rate of 60 watts, then the whole contraption is getting 60 watts of power from the wall outlet.

Power = (voltage) x (current)

60 watts = (120 v) x (current)

Current = (60 watts) / (120 v)

<em>Current = 0.5 Ampere</em>

5 0

3 years ago

Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po

N76 [4]

Answer:

The intensity of the light after passing through the two polarizing filters is 4.00 units

Explanation:

Io = 25 + A

= 25 + 9

= 34

Intensity of light through first polarizer,

I1 = Io/2

= 34/2

= 17 units

angle between axis of frist and second polarizers, theta = 55 + 6

= 61 degrees

Intensity of light passing through second polarizer,

I2 = I1*cos^2(theta)

= 17*cos^2(61)

= 4.00 units

6 0

3 years ago

34)You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then

velikii [3]

Answer:

$F_H_n=230.04 N$

The Required horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

$F_H_n=F_f\\F_h=mg*u$

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

$200=mg*u$ Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

$F_H=(1.42m*g*.81u)$ Eq (2)

Divide Eq(2) and Eq (1)

$\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N$

The Required horizontal force is 230.04N

4 0

3 years ago