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choli [55]
3 years ago
11

What is the electric power used by an appliance if the current is 2 A and the voltage is 50 volts? Show work and include all uni

ts
Physics
1 answer:
dem82 [27]3 years ago
3 0
Both the answers from "Big D" and "N9KXF" are correct, but not complete. There is one aspect that needs to be remembered for AC power - the phase relationship between the voltage and current. 

<span>For AC power, the more complete equation for power is P = V * I * cosx, where x is the phase angle between voltage and current. </span>

<span>If the phase angle between voltage and current is 90 degrees, then even though current may be flowing, the total power over a cycle would be zero. Reactive elements like inductors and capacitors have a 90degree phase relationship between voltage and current. They do not "dissipate" energy. They store energy and give it back during different parts of the AC cycle. </span>
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A roller coaster travels down a 120 m track in 12.5 seconds how fast does the roller coaster go
KIM [24]

Answer:

9.6m/s

Explanation:

Using the equation S=d/t where s=speed, d=distance, and t=time

plug in the known variables

S=120m/12.5s

S=9.6m/s

4 0
3 years ago
As a result of photosynthesis is the production of sugar molecules known as what?
Bezzdna [24]

Answer:

Glucose

Explanation:

4 0
3 years ago
Read 2 more answers
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
2 years ago
An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro
kupik [55]

Answer:

a) Q_{in} = 13.742\,kW, b) \Delta S = 370.15\,\frac{kJ}{K}

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

Q_{in} +U_{sys,1} - U_{sys,2} = 0

Q_{in} = U_{sys,2} - U_{sys,1}

Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})

Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)

Q_{in} = 13.742\,kW

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

\Delta S = \frac{Q_{in}}{T_{in}}

\Delta S = \frac{13.742\,kJ}{370.15\,K}

\Delta S = 370.15\,\frac{kJ}{K}

3 0
3 years ago
A lone neutron spontaneously decays into a proton plus an _______
frosja888 [35]

Answer:

A lone neutron spontaneously decays into a proton plus an electron.

Explanation:

In an atom, nuclei contain protons and neutrons, which are the fundamental particles of an atom. Neutrons are stable and uncharged particles inside a nucleus.

For 15 times during its lifetime, a free neutron decays and breaks down into more smaller particles.This breakdown causes problems in nuclear reactors, as they start decaying and emit radiations of different wavelengths.

A neutron undergoes the decaying process to produce an electron, a proton, and energy.

The reaction of neutron decay:

  n0 → p+ + e− + νe

5 0
3 years ago
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