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3 years ago

6

For each of the following, draw the resulting wave when the two pulses occupy the same space. Unless otherwise noted, each wave

has an amplitude of +/- A.

1 answer:

frozen [14]3 years ago

4 0

Answer:

A

Explanation:

because it is are of circle

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Which of the following would cause the force of kinetic friction to increase?

MrRa [10]

Answer:

c

Explanation:

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2 years ago

The rotating loop in an AC generator is a square l on each side. It is rotated at frequency f in a uniform field of B. The gener

spin [16.1K]

Answer:

Explanation:

Area of square loop = L²

Flux Φ = area x magnetic field

= L²B

Frequency = f

angular velocity ω = 2πf

a )

Let at time t = 0 , the magnetic field is making 90 degree with the face of the loop

flux through loop = L²B

After time t , coil will turn by angle ω t = 2πft

Flux through the loop = L²B cosω t

Φ (t) = L²B cosω t

= L²B cos2πft

b )

emf induced e

= - d/dt [Φ (t)]

= - d/dt [ L²B cosω t]

= L²B ω sinω t

= L²B 2πf sin2πft

c )

current = e / R

(L²B ω/ R ) sinω t

Power delivered

P(t) = VI ,

VOLT X CURRENT

= AB ω sinω t X ( AB ω/ R ) sinω t

= L⁴B² 4π²f²/R sin²2πft

e )

torque = MB sinω t

τ(t) = i(L²B ) sinω t

= (L²B ω/ R ) sinω t x (L²B ) sinω t

= (L²B )²ω/ R sin²ω t

= (L²B )² 2πf/ R sin²2πft

8 0

2 years ago

What happens when light strikes a translucent object??

Ivan

It allows the light to pass through but no other object can

4 0

2 years ago

Read 2 more answers

Which statements about our solar system are false?

scoray [572]

Answer:

D and F are false

Explanation:

A B and C are true.

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2 years ago

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c

FrozenT [24]

Answer:

a) $\Delta \dot K = 24.570\,kW$ , b) $\dot W_{out} = 12729.15\,kW$ , c) $A_{in} = 0.0136\,m^{2}$

Explanation:

A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

$-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2}) = 0$

a) Change in kinetic energy

$\Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})$

$\Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]$

$\Delta \dot K = 24570\,W$

$\Delta \dot K = 24.570\,kW$

b) Power output

$\dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})$

$\dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW$

$\dot W_{out} = 12729.15\,kW$

c) Turbine inlet area

Turbine inlet area can be found by using the following expressions:

$\dot V_{in} = \dot m \cdot \nu_{in}$

$\dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)$

$\dot V_{in} = 1.089\,\frac{m^{3}}{s}$

$A_{in} = \frac{\dot V_{in}}{v_{in}}$

$A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }$

$A_{in} = 0.0136\,m^{2}$

8 0

2 years ago