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Elina [12.6K]
3 years ago
9

A toaster with a power rating of 1,000 watts is plugged into a household voltage of 120 volts. How much current runs through the

toaster?
A) 120,000 amps
B) 880 amps
C) 8.3 amps
D) 0.120 amps
Physics
2 answers:
deff fn [24]3 years ago
6 0

Answer : I = 8.33 A

Explanation :

It is given that,

Power rating of the toaster, P = 1000 watts

Household voltage, V = 120 volts

We know the relation between power, voltage and the current is :

P=V\times I

I=\dfrac{P}{V}

I=\dfrac{1000\ W}{120\ V}

I = 8.33\ A

8.33 A of current runs through the toaster.

Hence, this is the required solution.

Lunna [17]3 years ago
4 0
C) 8.3 i believe i wouldnt take me 100%
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12
Svet_ta [14]

Answer:

See below ~

Explanation:

An object will sink in water when its density is greater than that of water, which is 1 g/cm³.

Volume of the box is <u>1331 cm³</u>. (11³)

Maximum mass of sand will be 1331 g. [because 1331/1331 = 1 g/cm³]

  • Volume of sand = Mass of sand / Density of sand
  • Volume (sand) = 1331/3.5
  • Volume (sand) = 380.29 cm³

If the volume of sand is <u>greater than 380.29 cm³</u>, the box will sink in water.

6 0
3 years ago
Two satellites are orbiting earth at different altitudes. Which satellite orbits at a higher speed v around earth?
aniked [119]

Answer:

Low satellite has high orbital velocity

Explanation:

let v be the orbital speed of the satellite orbiting at a height h is given by

v=\sqrt{\frac{GM}{R+h}}

where, M be the mass of planet, r be the radius of planet and h be the height of planet from the surface of planet.

here we observe that more be the height lesser be the orbital velocity.

So, a satellite which is at low height has high orbital velocity.

3 0
4 years ago
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV
Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

 The radius  of  the dish is R = 18cm

Explanation:

  From the question we are told that

     The radius of the orbit is  = R = 35,000km = 35,000 *10^3 m

    The power output of the power is  P = 1 kW = 1000W

   The electric vector amplitude is given as E = 0.1 mV/m = 0.1 *10^{-3}V/m

    The area of thereciever  is   A_R = 5cm^2

Generally the intensity of the dish is mathematically represented as

         I = \frac{P}{A}

Where A is the area orbit which is a sphere so this is obtained as

          A = 4 \pi r^2

              = (4 * 3.142 * (35,000 *10^3)^2)

              =1.5395*10^{16} m^2

  Then substituting into the equation for intensity

          I_s  =  \frac{1000}{1.5395*10^{16}}

            = 6.5*10^ {-14}W/m2

 Now the intensity received by the dish can be mathematically evaluated as

              I_d = \frac{1}{2}  * c \epsilon_o E_D ^2

  Where c is thesped of light with a constant value  c = 3.0*10^8 m/s

              \epsilon_o is the permitivity of free space  with a value  8.85*10^{-12} N/m

              E_D is the electric filed on the dish

So  since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

      Now making the eletric field intensity the subject of the formula

                  E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }

substituting values

                 E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }

                       = 7*10^{-6} V/m

The incident power on the dish is what is been reflected to the receiver

                P_D = P_R

Where P_D is the power incident on the dish which is mathematically represented as

              P_D = I_d A_d

                   = \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2)

And  P_R is the power incident on the dish which is mathematically represented as

                 P_R = I_R A_R

                       = \frac{1}{2} c \epsilon_o E_R^2 A_R

Now equating the two

                \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2) =  \frac{1}{2} c \epsilon_o E_R^2 A_R

   Making R the subject we have

                   R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }

Substituting values

                   R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }

                     R = 18cm

8 0
3 years ago
An exothermic reaction is taking place in a test tube what would you expect to feel when you touch the outside of the test tube
ValentinkaMS [17]

Answer:

Heat, Let me explain,

Explanation:

An exothermic reaction is like... Lighting a candle so it produces  a small amount of heat in less than a second and then normal heat.

7 0
4 years ago
If the accepted value of a wave is 121 m/s, who has the most accurate method of measuring the speed of a wave?
qaws [65]
The answer would be erin out of all of them thank me later :)
5 0
3 years ago
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