How is energy wasted by the radio

Inside a freely falling runaway elevator, your

nasty-shy [4]

The answer is apparent weight is zero.

You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).

You still are being affected by the Earth's gravity.

Not all because of the previous two statements.

Not none because apparent weight is zero as you are falling.

7 0

3 years ago

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The energy generated by the sun travels to Earth as electromagnetic waves of varying lengths. Which statement describes an elect

Paha777 [63]

Answer: 4(45)

Explanation:

It’s not equal

3 0

2 years ago

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Work is done on a locked door that remains closed while you try to pull it open. True False.

dsp73

Answer:False

Explanation:

Work is being done on a body when it causes displacement of body on the application of force

$Work\ done=Force\times displacement$

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.

Thus the above-given statement is false

8 0

3 years ago

The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on

sammy [17]

Answer: First, we determine the circumference of the Mars by the equation below.

C = 2πr

Substituting the known values,

C = 2(π)(3,397 km) = 6794π km

To determine the tangential speed, we divide the circumference calculated above by the time it takes for Mars to complete one rotation and that is,

tangential speed = 6794π km / 24.6 hours = 867.64 km/h

6 0

2 years ago

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If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

--

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

--

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

--

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago