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bagirrra123 [75]
2 years ago
7

Drag each label to the correct location on the table. sort the processes based on the type of energy transfer they involve. cond

ensation freezing deposition sublimation evaporation melting reset next
Physics
1 answer:
NemiM [27]2 years ago
4 0

Explanation:

Xhxdyf is a greatg resource for all 6 ge that can be 300 and you can get in a on line with your favorite own web and web site to help you find a great place to work for your free online dating site that you will enjoy and use the

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An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether
Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

I = \frac{P}{A}

where :

\frac{I}{c}= \frac{F}{A}

Replacing that into the above previous equation; we have:

\frac{P}{Ac}=\frac{F}{A}

F = \frac{P}{c}

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the  astronaut and a = its acceleration

we as well say;

\frac{P}{c} = ma

a = \frac{P}{mc}

Replacing P with 1000 W ; m with 80 kg and 3*10^{8} \  m/s for  c

Then; a = \frac{1000 \ W}{(80)(3.0*10^8)}

a = 4.2*10^{-8} \ m/s

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2

\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

v_1 = at_1

where ;

v_1 = final \ velocity

replacing t_1 = 1.0 \ h and a =  4.2*10^{-8} \ m/s; Then:

v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})

v_1 =  1.51 *10^{-4} \ m/s

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

\Delta x_2 = d - \Delta x_1

where;

\Delta x_2 = the final distance

d = total distance

So;

\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m

The time taken to reach the final distance can be calculated as:

t_2 = \frac{\Delta x_2 }{v_1}

where;

t_2 = is the  time to  reach the final distance

Replacing 4.73 for {\Delta x_2 } and  1.51*10^{-4} m/s for v_1

t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}

t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )

t_2 = 8.8 \ h

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to  reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0
3 years ago
17. Calculate the amount of gravitational potential energy at the top of one 4 points hill. The mass of the coaster is 500 kg. T
MissTica

Answer: D. 292,338 J

This is the correct answer :)

4 0
3 years ago
What is one way to lower gravitational potential energy?
il63 [147K]

Answer:

decrease the height

Explanation:

height is directly proportional to the g.p.e

4 0
3 years ago
If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m
algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

H=ut+\dfrac{1}{2}at^2

Here, a = g

H=\dfrac{1}{2}gt^2            

H=\dfrac{1}{2}\times 9.8\times (2.261)^2

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

7 0
3 years ago
1. Which segment on this position vs time graph represents the object returning back to the origin?
Monica [59]

9514 1404 393

Answer:

  D. Segment D

Explanation:

The origin is represented on the graph by the position (y-value) being 0. The portion of the graph that shows position being non-zero and ending at 0 is the portion marked <em>segment D</em>.

5 0
3 years ago
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