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Arada [10]
2 years ago
8

What is the voltage drop across the 10.0 2 resistor?

Physics
1 answer:
amid [387]2 years ago
4 0

Answer:

the answer is equal to 2.00v

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A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
Find the current flowing out of the battery.​
klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0
3 years ago
If you stood on a planet with four times the mass of Earth, and twice Earth's radius, how much would you weigh?
nikdorinn [45]

Answer:

1/4 times your earth's weight

Explanation:

assuming the Mass of earth = M

Radius of earth = R

∴ the mass of the planet= 4M

the radius of the planet = 4R

gravitational force of earth is given as = \frac{GM}{R^{2} }

where G is the gravitational constant

Gravitational force of the planet = \frac{G4M}{(4R)^{2} }

                                                       =\frac{G4M}{16R^{2} }

                                                       =\frac{GM}{4R^{2} }

recall, gravitational force of earth is given as = \frac{GM}{R^{2} }

∴Gravitational force of planet = 1/4 times the gravitational force of the earth

you would weigh 1/4 times your earth's weight

3 0
3 years ago
Joe the house painter stands on a uniform oak board weighing 600 N and held up by vertical ropes at each end. Joe has been dieti
Mamont248 [21]

Answer

given,

weight of the oak board = 600 N

Weight of Joe = 844 N

length of board = 4 m

Joe is standing at 1 m from left side

vertical wire is supporting at the end.

Assuming the system is in equilibrium

T₁ and T₂ be the tension at the ends of the wire

equating all the vertical force

T₁ + T₂ = 600 + 844

 T₁ + T₂ = 1444...........(1)

taking moment about T₂

 T₁ x 4 - 844 x 3 - 600 x 2 = 0

 T₁ x 4 = 3732

 T₁ = 933 N

from equation (1)

 T₂ = 1444 - 933

 T₂ = 511 N

3 0
3 years ago
From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:
Oduvanchick [21]

a)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m


also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec


b)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

3 0
3 years ago
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