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poizon [28]
1 year ago
10

A 3. 5-a current is maintained in a simple circuit with a total resistance of 1500 ω. What net charge passes through any point i

n the circuit during a thirty second interval?.
Physics
1 answer:
melamori03 [73]1 year ago
7 0

Electric Current:

Electric current is the flow of charge through a given circuit per unit time. Electric current is one of the components needed to calculate the electric power that a device needs to operate and do work. Electric current is measured in amperes (A), which is equal to:

            1A = 1 C/ s

Recall that the coulomb (C) is the unit for charge while the second (s) is the unit for time

Given: I = 3.5

A is the current

Δt =30 s is the time interval

                     A =ΔQ/ΔT

Net charge = 100C

Electricity is produced when an electric current runs through a circuit.

How does electric current work?

A current of electricity is a steady flow of electrons. When electrons move from one place to another, round a circuit, they carry electrical energy from place to place like marching ants carrying leaves. Instead of carrying leaves, electrons carry a tiny amount of electric charge.

Learn more about Electric current :

brainly.com/question/27003377

#SPJ4

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The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril
STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

                         T = 5252*P / N

                         T = 5252*5.85 / 1900

                         T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

                          T = F*r

Where,   r = d / 2

                          F = 2T / d

                          F = 2*16.171 / 0.98

                          F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

4 0
3 years ago
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A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no apprecia
Elenna [48]

Answer:A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no ..

6 0
2 years ago
A 20.5 kg ball moving at 38.5 m/s on a horizontal, frictionless surface runs into a light spring of force
tekilochka [14]

Answer:

7.68 m

Explanation:

Kinetic energy in ball = elastic energy in spring

KE = EE

½ mv² = ½ kx²

mv² = kx²

x = v √(m / k)

x = (38.5 m/s) √(20.5 kg / 515 N/m)

x = 7.68 m

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3 years ago
Una banda transportadora se utiliza para cargar canastos en un avión de carga, si µs = 0.5. ¿Cuál es el ángulo máximo de elevaci
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3 years ago
The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
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Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

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