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Mkey [24]
2 years ago
7

What is the resistance (in Ω) of thirty-seven 330 Ω resistors connected in series? 12210 Correct: Your answer is correct. Ω (b)

What is the resistance (in Ω) of thirty-seven 330 Ω resistors connected in parallel?
Physics
1 answer:
notsponge [240]2 years ago
7 0

Answer:

(a) 12210  Ω

(b) 9.92 Ω

Explanation:

From Electricity,

(a)

When resistors are connected in series, the total resistance is the sum of the individual resistance.

From the question,

Rt = R1 + R2 + R3................................. R37 ....................... Equation 1

Where Rt = Total resistance of combined resistance, R1 = resistance of the first resistor, R2 = Resistance of the second resistor .......... and so on up to the 37th resistor.

Note: Since The resistor are identical.

Rt = n(R) ..................... Equation 2

Where n = number of resistor, R = resistance of each resistor.

Given: n = 37, R = 330 Ω

Substitute into equation 2

Rt = 37(330)

Rt = 12210 Ω

(b)

For parallel connection,

Rt = R/n...................... Equation 2

Given: R = 330 Ω, n = 37

Substitute into equation 2

Rt = 330/37

Rt = 8.92 Ω

Hence the resistance when connected in parallel = 9.92 Ω

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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

t=time within  1 deg C of it steady state is 8.31s

Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

7 0
3 years ago
Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J
olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= \sqrt{(8)^2+(-2)^2}

|v|= \sqrt{68} = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0
3 years ago
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr
mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × 10^{-19} = 1/2 × 1.672 × 10^{-27} ×v²

v = 3.5 × 10^{7} m/s

so

radius will be

radius = \frac{m*v}{B*q}   ........2

radius =  \frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}  

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

8 0
3 years ago
Read 2 more answers
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci
Alecsey [184]

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

Answer:  

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

8 0
3 years ago
An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an unk
devlian [24]

Answer:

3.192 m/s

Explanation:

t = Time taken = 0.900 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 1.1 meters

a = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1.1-\frac{1}{2}\times 9.81\times 0.9^2}{0.9}\\\Rightarrow u=-3.192\ m/s

Velocity of the elevator when it snapped is 3.192 m/s

4 0
3 years ago
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