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rusak2 [61]
3 years ago
6

A standard carbon resistor has a gold band to indicate + 5% tolerance. If its resistance is 3,500 , what are the upper and lower

limits for its resistance? OA . 3495 - 3505 2 OB. 3300 Q - 3600 0 OC. 3325 N - 3675 OD 3450 - 35500​
Engineering
1 answer:
Lapatulllka [165]3 years ago
7 0

Answer:

  C.  3325 Ω - 3675 Ω

Explanation:

5% of 3500 Ω is ...

  0.05 × 3500 = 175

The lower limit is this amount less than the nominal value:

  3500 -175 = 3325

The upper limit is the nominal value plus the tolerance:

  3500 +175 = 3675

The lower and upper limits are 3325 Ω and 3675 Ω, respectively.

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Answer is c Heisenberg's uncertainty principle

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\Delta x\times \Delta \overrightarrow{p}\geq \frac{h}{4\pi }

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A heat engine operates between 2 reservoirs at TH and 18oC. The heat engine receives 17,000 kJ/h from the high temperature reser
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3 years ago
Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h
eimsori [14]

Answer:

t = 6179.1 s = 102.9 min = 1.7 h

Explanation:

The energy provided by the resistance heater must be equal to the energy required to boil the water:

E = ΔQ

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where.

η = efficiency = 84.5 % = 0.845

P = Power = 2.61 KW = 2610 W

t = time = ?

m = mass of water = 6.03 kg

H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Therefore,

(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

t = \frac{1.362\ x\ 10^7\ J}{2205.45\ W}

<u>t = 6179.1 s = 102.9 min = 1.7 h</u>

4 0
3 years ago
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
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Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
3 years ago
Acertain foundation will experience a bearing capacity failurewhen it is subjected to a downward load of 2200 kN. Using ASD with
ehidna [41]

Answer:

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Explanation:

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