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vladimir1956 [14]

2 years ago

12

To heat the house, the boiler transfers 15 MJ of energy in 10 minutes.

1 answer:

Harrizon [31]2 years ago

8 0

The formula is P = E/t, where P means power in watts, E means energy j , and t means time in seconds. This formula states that power is the consumption of energy per unit of time.

P = 15 M / 10*60

M = mega = 10⁶

15 *10⁶ / 600

= 25000 watt

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In creating his definition of horsepower, James Watt, the inventor of the steam engine, calculated the power output of a horse o

Studentka2010 [4]

Answer:

Part a)

$F = 182.3 Lb$

Part b)

$P = 0.55 HP$

Explanation:

Diameter of the circle = 24 ft

Diameter = 731.52 cm = 7.3152 m

now the horse complete 144 trips in one hour

so time to complete one trip is given as

$t = \frac{3600}{144} s$

$t = 25 s$

now the speed of the horse is given as

$v = \frac{2\pi r}{t}$

$v = \frac{\pi(7.3152)}{25}$

$v = 0.92 m/s$

Part a)

Now we know that the power is defined as rate of work done

it is given as

$P = F v$

$746 = F(0.92)$

$F = 810.9 N$

$F = 182.3 Lb$

Part b)

Work done to climb up to 3 m height is given by

$W = mgh$

now we have

$Power = \frac{Work}{time}$

$P = \frac{mgh}{t}$

$P = \frac{(70kg)(9.81)(3)}{5.0s}$

$P = 412.02 Watt$

now we know that 1 HP = 746 Watt

so we have

$P = \frac{412}{746} = 0.55 HP$

8 0

3 years ago

Whats is the resistance of a bulb with a current of 60a in a circuit with a 12v battery

lara [203]

Answer:

60a in a circuit with a 12v battery

Explanation:

60a in a circuit with a 12v battery

6 0

2 years ago

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

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Unless if all forces cancel each other out , the object will no longer be in equilibrium

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The answer would be
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