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expeople1 [14]
3 years ago
5

The diagram below shows a light ray striking Medium A and Medium B at the same angle.

Physics
1 answer:
zepelin [54]3 years ago
6 0

Answer: D  Medium B is denser than Medium A because light bends more in Medium B.

Explanation: The more the light bends the less energy it has

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Answer:

Larger tubines generate more electricity.

Explanation:

Larger blades allow the turbine to capture more of the kinetic energy of the wind by moving more air through the rotors. However, larger blades require more space and higher wind speeds to operate. This distance is necessary to avoid interference between turbines, which decreases the power output.

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1. a) Name 2 properties of waves that change if the wave changes media.
Drupady [299]
1.) The properties of a wave are the following.
a) <span><span>Amplitude - the height of the wave, measured in meters.
</span><span>b) Wavelength - the distance between adjacent crests, measured in meters.
</span><span>c) Period - the time it takes for one complete wave to pass a given point, measured in seconds.
</span><span>d) Frequency - the number of complete waves that pass a point in one second, measured in inverse seconds, or Hertz (Hz).
</span><span>e) Speed - the horizontal speed of a point on a wave as it propagates, measured in meters / second.

Among these properties, PERIOD and SPEED changes if the wave changes media. The others remain the same.

2. Speed = Wavelength * Frequency
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Which two forms of energy does a hairdryer convert electric energy into? A.)chemical energy
Ne4ueva [31]
D, because the hairdryer uses a fan
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4 years ago
Read 2 more answers
How is gravity simulated on the hermes spacecraft?
gulaghasi [49]

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3 0
3 years ago
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la
Nookie1986 [14]

Answer:

h=53.09m         (2)

v_{min}>5.05m/s

v_{max}

Explanation:

<u>a)Kinematics equation for the first ball:</u>

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h       initial position is the building height

v_{o}=8.9m/s      

The ball reaches the ground, y=0, at t=t1:

0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}

h=1/2*g*t_{1}^{2}-v_{o}t_{1}           (1)

Kinematics equation for the second ball:

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h       initial position is the building height

v_{o}=0       the ball is dropped

The ball reaches the ground, y=0, at t=t2:

0=h-1/2*g*t_{2}^{2}

h=1/2*g*t_{2}^{2}         (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03              (3)

We solve the equations (1) (2) (3):

1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}

g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)

g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)

-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)

2.06*gt_{1}-2v_{o}t_{1}=g*1.06

t_{1}=g*1.06/(2.06*g-2v_{o})

vo=8.9m/s

t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s

t2=t1-1.03              (3)

t2=3.29sg

h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m         (2)

b)t_{1}=g*1.06/(2.06*g-2v_{o})

t1 must :   t1>1.03  and t1>0

limit case: t1>1.03:

1.03>9.81*1.06/(2.06*g-2v_{o})

1.03*(2.06*9.81-2v_{o})

20.8-2.06v_{o}

(20.8-10.4)/2.06

v_{min}>5.05m/s

limit case: t1>0:

g*1.06/(2.06*g-2v_{o})>0

2.06*g-2v_{o}>0

v_{o}

v_{max}

8 0
4 years ago
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