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Hoochie
23 hours ago
12

While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali

ng with circuits though, it is often more useful to know the rate at which energy is changing. As you may recall, this quantity is also known as power.Consider a resistor with resistance R. The current through the resistor is I, and the voltage across the resistor is V. Which of these formulae can be used to find the power drawn by the conductor?
Physics
1 answer:
serg [2.5K]23 hours ago
7 0
The question lacks completeness. The options given include RI^2, I^2/R, R/I^2, R/V^2, RV^2, V^2/R, VI, and VIR. The applicable equations are P = RI^2, P = V^2/R, and P = VI, as they all pertain to calculating power in a circuit context.
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Compare the time period of two simple pendulums of length 4m and 16m at a place.
Ostrovityanka [2204]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

5 0
23 days ago
Ceres, Pluto, and Eris are all round in shape and classified as:_________ A) Leftover planetesimals that formed inside the frost
Keith_Richards [2256]

Answer

Ceres, Pluto, and Eris are categorized as DWARF PLANETS.

A) Remaining planetesimals formed within the frost line are referred to as ASTEROIDS.

B) METEORITES are fragments of asteroids that have landed on Earth.

C) COMETS are celestial objects that are often visible with their long tails.

D) COMETS are also planetesimals that were left over and originated in the region of the solar system dominated by the jovian planets.

E) Meteor showers are linked to debris from COMETS.

5 0
21 day ago
Most calculators operate on 6.0 V. If, instead of using batteries, you obtain 6.0 V from a transformer plugged into 110-V house
kicyunya [2264]
1/0.0545. The transformation ratio of primary coil turns to secondary coil turns is directly proportional to the voltage transformation occurring. With 6.0 V on the secondary side (output) and 110 V on the primary side (input), the voltage ratio is calculated as 6/110 = 0.0545. This means for each turn in the primary coil, there are 0.0545 turns in the secondary coil.
6 0
11 days ago
A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the base at a velocity of 15.5 m/s
Yuliya22 [2420]
This can be determined using the principle of energy conservation. The ski lift begins with a velocity of v= 15.5 m/s, and all of its kinetic energy Ek converts into potential energy Ep, thus we set Ep equal to Ek.
Because Ek is given by (1/2)*m*v², where m denotes mass and v represents speed, while Ep equals m*g*h, where m is mass, g is 9.81 m/s², and h is height. Now:

Ek=Ep 

(1/2)*m*v²=m*g*h, canceling out the mass,

(1/2)*v²=g*h, rearranging for height by dividing by g,

(1/2*g)*v²=h and substituting the values:

h=12.245 m. The hill's height rounded to the nearest tenth is h=12.25 m.

5 0
18 days ago
Read 2 more answers
A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote
Keith_Richards [2256]
1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
U=q V
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is q=4.0 \mu C = 4.0 \cdot 10^{-6}C, and the potential energy is U=5.2 \cdot 10^{-3} J, thus we may rearrange the earlier formula to find the electric potential at the tip:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V

2) Using this same formula, if the charge changes to q=2.0 \mu C = 2.0 \cdot 10^{-6} C, the resulting electric potential will be:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V
8 0
10 days ago
Read 2 more answers
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