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ira [324]
4 years ago
12

A 240-volt, 2-amp motor is connected to a three-wire, 120/240-volt system. Connected between the black wire and neutral are four

200-watt, 120-volt lamps and a 120-volt, 1-amp motor. Between the red wire and neutral are three 200-watt, 120-volt lamps, one 1.67-amp motor and one 120-volt, 1-amp motor. (Round the FINAL answer to two decimal places.). How many amps flow in the red wire?

Physics
1 answer:
pantera1 [17]4 years ago
6 0

Answer:

(i)The current flow in black wire = 9.67 A (ii) The current low in the red wire is 9.68 A (iii) The current flow in neutral wire is  15.36 A (iv) when 240 volt were disconnected current  in black wire is 7.68 A (v) when 240 volt were disconnected current in red wire is 7.68 A (vi) 15.36 A (vii) 6.34 (viii) 9.68 A (ix) 12.02 A

Explanation:

Solution

The current drawn by one amp is

I =P/V

I =200/120

I= 1.67 A

(i) The current flow in the black wire  is

IBK = 4 * 1.67 A + 1A + 2A

IBK = 9.67 A

(ii) Current flow in the red wire is

IRD = 3 * 1.67 A + 1.67 A + 1A + 2A

= 8.68A + 1 A = 9.68 A

Note: Kindly find an attached copy of part of the solution to the given question above.

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The chart show the masses and velocities of two colliding objects that stick together after a collision.
finlep [7]

Answer:

<u><em></em></u>

  • <u><em>1,500 kg.m/s</em></u>

Explanation:

First, arrange the information in a table:

Object        Mass (kg)           Velocity (m/s)

  A                    200                      15

  B                     150                    - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

<u>Momentum before the collision, P₁</u>:

          P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}

         P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\  P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s

<u>Momentum after the collision</u>:

  • As stated, it es equal to the momentum before the collision: 1,500 kg . m/s
6 0
4 years ago
A negatively charged particle is attracted to
Tatiana [17]

Answer: I think the answer is D.

Positively charged particles.

8 0
3 years ago
A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected
vekshin1

Answer:

(A) 9.14\times 10^{-9}sec

(B) 6.20\times 10^{-3}A

Explanation:

We have given inductance L=5.41\mu H=5.41\times 10^{-6}H

Resistance R=0.949kohm=0.949\times 10^3ohm

Time constant of RL circuit is equal to \tau =\frac{L}{R}

\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

i=i_0(1-e^{\frac{-t}{\tau }})

According to question

0.799i_0=i_0(1-e^{\frac{-t}{\tau }})

e^{\frac{-t}{\tau }}=0.201

{\frac{-t}{\tau }}=ln0.201

{\frac{-t}{5.7\times 10^{-9} }}=-1.6044

t=9.14\times 10^{-9}sec

(b) Current at t=\tau sec

i=i_0(1-e^{\frac{-t}{\tau }})

i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})

i=6.20\times 10^{-3}A

3 0
3 years ago
Given that a fluid at 260°F has a kinematic viscosity of 145 mm^2/s, determine its kinematic viscosity in SUS at 260°F.
Oduvanchick [21]

Answer:

kinematic viscosity in SUS is = 671.64 SUS

Explanation:

given data

kinetic viscosity = 145 mm^2/s

we know

1 mm = 0.1 cm

so kinetic viscosity in cm is \nu =145 (0.1)^{2} =1.45 cm^{2}/s

other unit of kinetic viscosity is centistokes

1 cm^{2}/s = 100 cst

so 1.45 cm^2/s will be 145 cst

if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS

5 0
4 years ago
The latent heat of fusion of alcohol is 50 kcal/kg and its melting point is -54oC. It has a specific heat of 0.60 in its liquid
sashaice [31]

Answer:

C

Explanation:

To melt the alcohol

Heat needed = M . L  = 2 . 25  = 50 kcal

To warm up the alcohol

Heat needed = M . sp. ht. . ∆t   = 2 . 0.6 . 100  = 120 kcal

Total heat needed = 170 kcal

Assuming that 0.6 kcal/ kg / ˚C  is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)

5 0
3 years ago
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