All categories

2 years ago

A 75.0 W bulb is connected to a

Physics

1 answer:

adoni [48]2 years ago

6 0

Answer:

Explanation:

W = 75 watts

V = 110 volts

Formula

W = V * I

Solution

75 = 110 * I Divide by 110

75 / 110 = I

I = 0.6818 Amperes

You might be interested in

A ship tows a submerged cylinder, which is 1.5 m in diameter and 22 m long, at 5 m/s in fresh water at 208C. Estimate the towing

Scilla [17]

Answer: a) P = 120kw

b) P = 800kw

Explanation: Please find the attached file for the solution

8 0

3 years ago

Read 2 more answers

An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the

zhannawk [14.2K]

B is the answer to this truly did the math

3 0

3 years ago

Read 2 more answers

If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in

Elza [17]

Answer:

The reflected resistance in the primary winding is 6250 Ω

Explanation:

Given;

number of turns in the primary winding, $N_P$ = 50 turns

number of turns in the secondary winding, $N_S$ = 10 turns

the secondary load resistance, $R_S$ = 250 Ω

Determine the turns ratio;

$K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5$

Now, determine the reflected resistance in the primary winding;

$\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms$

Therefore, the reflected resistance in the primary winding is 6250 Ω

6 0

3 years ago

PLEASE HELP!!!

Arisa [49]

Answer:

Each type of electromagnetic radiation have been matched to their descriptions as follows:

Ultraviolet light A) Given off by very hot objects, such as the sun

Gamma rays B) Given off by radioactive substances

Microwaves C) Used in radar and to heat food

X‒rays D) Used in the medical field to see bone structure

Explanation:

PLS GIVE BRAINLEST

6 0

3 years ago

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6

lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

The amplitude of induced emf by this signal between the ends of the receiving antenna is

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

Here,

I = Current

$\mu_0$ = Permeability at free space

c = Light speed

d = Distance

Replacing,

$\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}$

$\epsilon_{max} = 0.05434V$

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0

4 years ago