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Savatey [412]
3 years ago
9

how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?

Physics
1 answer:
ExtremeBDS [4]3 years ago
8 0
3 Amper 9 divided by 3
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Hello....I need help with this question.
musickatia [10]

Answer:

3.2 m/s²

Explanation:

Acceleration can be calculated as:

v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)

25 m/s  = 9 m/s  + a(5 s)  (a is unknown)

16 m/s  = a(5 s)

a = 3.2 m/s²

We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).

6 0
3 years ago
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Which process is involved in the formation of a galaxy
zubka84 [21]

Answer:

SUN

Explanation:

EARTH

8 0
3 years ago
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The time required for one cycle, a complete motion that returns to its starting point, is called the _____.
GenaCL600 [577]

Answer: the Time that is required is called a period

8 0
3 years ago
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

M = \frac{gr^2}{G}

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

g = 9.8m/s^2\\G  = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m

Replacing at this equation we have that

M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg

The Volume of a Sphere is equal to

V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3

Therefore using the relation between mass, volume and density we have that

\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3

6 0
3 years ago
A coil is connected in series with a 12.6 kΩ resistor. An ideal 65.2 V battery is applied across the two devices, and the curren
ivolga24 [154]

To solve the problem, start by applying the concepts related to current in an RL circuit. The current is defined exponentially and using Ohm's law we can put the initial current in terms of the voltage and resistance. Consecutively with the calculated time constant we can find the respective inductance. For the second part we will apply the electrical potential energy connectors to find the amount of stored energy.

PART A)

i = i_0 (1-e^{-t/T})

i = (V/R)(1-e^{-t/T})

0.00402 = (65.2/12600)(1-e^{-0.00402/T})

T = 0.002679s

T = L/R

Inductance can be defined then,

L = RT

L = (12600)(0.002679)

L = 33.75H

PART B) Now the energy is given under the terms:

E = \frac{1}{2}Li^2

E = \frac{1}{2} (33.75)(0.00402)^2

E = 0.0002727J

Therefore the energy stored in the coil at this same moment is 0.0002727J

3 0
3 years ago
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