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2 years ago

What is the potential difference needed to achieve a current of 16A with a resistance of

Physics

1 answer:

Fiesta28 [93]2 years ago

6 0

Hi there!

We can use the following equation:
$\large\boxed{V = I R}$

V = potential difference (? V)
I = Current (16A)
R = Resistance (192Ω)

Plug in the givens and solve:

$F = 16 \times 192 = \boxed{3072 V}$

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What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.

Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

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3 years ago

A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height

krok68 [10]

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

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3 years ago

What are the four system of measurement

Rudiy27

Answer:

ask google

Explanation:

5 0

3 years ago

Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The

AnnyKZ [126]

Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y: opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N directed down

for equilibrium system

Fy= Ty=107.2 N

Fy=107.2 N upward component of the force acting on the stake

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3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

5 0

3 years ago