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2 years ago

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The light beacon of a lighthouse is situated a distance 2.00 km from a long, straight shoreline (shortest distance to the shorel

ine). The light rotates at a rate of 3.00 rev/minute. At a certain time, the illuminated spot on the shoreline is 4.00 km from the lighthouse. How fast is the illuminated spot moving at that point?

1 answer:

kolbaska11 [484]2 years ago

6 0

In triangle ABC

tanθ = BC/AB

tanθ = x/2

taking derivative both side relative to "t"

sec²θ (dθ/dt) = (0.5) (dx/dt)

sec²θ w = (0.5) v eq-1

when x = 4 ,

tanθ = 4/2

θ = tan⁻¹ (2) = 63.4 deg

using eq-1

sec²θ w = (0.5) v

sec²63.4 (3) (2pi/60) = (0.5) v

v = 3.13 m/s

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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

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3 years ago

Derive the kinetic equations for Vmax and KM using the transit time and net rate constant method as discussed in class and follo

Pepsi [2]

Answer:

Rate = vmax k3/k2+k3

Explanation:

The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, is referred to as Vmax.

This is usually expressed as the Km ie. Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax.

Please kindly check attachment for the step by step solution of the given problem.

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2 years ago

A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhauste

marusya05 [52]

516.154 megawatts of heat are <em>exhausted</em> to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river ( $Q_{out}$ ), in megawatts:

$Q_{out} = Q_{in} - W$

$Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W$ (1)

Where:

$\eta$ - Efficiency.
$W$ - Electric power, in megawatts.

If we know that $\eta = 0.39$ and $W = 330\,MW$ , then the energy rate exhausted to the river is:

$Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)$

$Q_{out} = 516.154\,MW$

516.154 megawatts of heat are <em>exhausted</em> to the river that cools the plant.

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7 0

2 years ago

A baseball has a mass of 0.15 kg and radius 3.7 cm. In a baseball game, a pitcher throws the ball with a substantial spin so tha

mylen [45]

Answer:

Rotational kinetic energy = 0.099 J

Translational kinetic energy = 200 J

The moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ .

Explanation:

Rotational kinetic energy is given by

$\text{RKE} = \frac{1}{2}I\omega^2$

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.

For a solid sphere,

$I = \frac{2}{5}mr^2$

where <em>m</em> is its mass and <em>r</em> is its radius.

From the question,

<em>ω</em> = 49 rad/s

<em>m</em> = 0.15 kg

<em>r</em> = 3.7 cm = 0.037 m

$\text{RKE} = \frac{1}{2}\times \frac{2}{5} mr^2\omega^2 = \frac{1}{5} mr^2\omega^2$

$\text{RKE} = \frac{1}{5} (0.15\ \text{kg})(0.037\ \text{m})^2(49\ \text{rad/s})^2 = 0.099\text{ J}$

Translational kinetic energy is given by

$\text{TKE} = \frac{1}{2} mv^2$

where <em>v</em> is the linear speed.

$\text{TKE} = \frac{1}{2} (0.15\ \text{kg})(52\ \text{m/s})^2 = 200\text{ J}$

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2 years ago

Read 2 more answers

a large truck has a mass of 20000kg. it is traveling at 28m/s along a staright road . calculate the kinetic energy

Semmy [17]

Answer:

The answer is

<h2>7,840,000 J</h2>

Explanation:

The kinetic energy KE of an object given it's mass and velocity can be found by using the formula

$KE = \frac{1}{2} m {v}^{2} \\$

where

m is the mass

v is the velocity

From the question

m = 20000kg

v = 28 m/s

It's kinetic energy is

$KE = \frac{1}{2} \times 20000 \times {28}^{2} \\ = 10000 \times 784$

We have the final answer as

<h3>7,840,000 J</h3>

Hope this helps you

3 0

3 years ago