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Alinara [238K]
3 years ago
8

Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr

oximate voltage drop will he experience? A. 1.8 V B. 2.8 V C. 3.5 V D. 5.9 V
Physics
1 answer:
castortr0y [4]3 years ago
8 0

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

R = \frac{2.525}{1000}\times 50

R = 0.126 ohm

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

V = iR

V = 11 \times 0.126

V = 1.4 Volts

so most appropriate answer in given options is

A. 1.8 Volts

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The speed limit sign in indiana reads 60 miles per hour. convert these units to meters per second
cluponka [151]

Answer:

26.822 m/s

Explanation:

60 mi/hr  *  5280 ft/mile *  1 hr / 3600 sec  *  12 in / foot * 1 meter / 39.37 in = <u>26.822 m/s</u>

3 0
2 years ago
Which statement is TRUE?
Georgia [21]
The answer is B I know this because I’m Asian
4 0
3 years ago
The flaming gorge bridge, in wyoming rises above a dry gulch. If you throw a rock straight out from the bridge, horizontally, an
Novosadov [1.4K]

Answer:

12.495m/s

Explanation:

Horizontal displacement is the range of the projectile motion.

The range is expressed as;

R = 2U/g

U is the speed at which the rock is thrown (initial speed)

g is the acceleration due to gravity.

Given

R = 255cm = 2.55m

g = 9.8m/s²

Required

Speed U

Substitute the given parameters into the formula as shown;

2.55 = 2U/9.8

Cross multiply

2U = 2.55×9.8

2U = 24.99

U = 24.99/2

U = 12.495m/s

Hence the speed that you thew the rock is 12.495m/s

7 0
3 years ago
A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s
SCORPION-xisa [38]

Answer:

a )  4.5 N.s

b) V =5 m/s    

Explanation:

given,

mass of rifle(M)  = 0.9 kg

mass of bullet(m)  = 6 g = 0.006 kg

velocity of the bullet(v)  = 750 m/s

a) momentum of bullet = m × v

                                  = 750 × 0.006

                                  = 4.5 N.s

b) recoil velocity                                                      

m × u + M × U = m × v + M × V

0  + 0  = 0.006 × 750 -  0.9 × V

V = \dfrac{4.5}{0.9}

V =5 m/s                    

 

5 0
3 years ago
Assuming a typical efficiency for energy use by the body, how many slices of pizza must you eat to walk for 2.5 h at a speed of
larisa86 [58]

Answer:

2.7 Pizzas.

Explanation:

The power required to walk through 5km in 1 hour is 380W.

A watt is basically Jules per second, then we need to standardized this measurement to second.

5km/hr is equal to,

\frac{5km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=1.389m/s

Walking by 2.5 hours is equal to a distance of,

d=v*t=1.389*(2.5*3600) = 12500m

The total energy required then would be,

E = \frac{380J}{1.389m/s}(12500)=3.4199*10^6J

Then we know that one pizza slice gives 1260*10^3J of energy, the total pizza needed are,

\eta = \frac{3.4199*10^6}{1260*10^3} = 2.7142

<em>Then you need to buy 3 pizza.</em>

6 0
4 years ago
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