<span>The total amount of kinetic energy and potential energy within a system is called "Mechanical Energy"

In short, Your Answer would be Option C

Hope this helps!</span>

6 0

3 years ago

A mountain slope and a skier make up a closed system with 21,800 J J of energy. When the skier starts her descent down the mount

belka [17]

Answer: 1,270 J

Explanation: APEX

4 0

3 years ago

Would the solar panel work under a fluorescent or halogen light? explain your response being sure to relate your observations to

lina2011 [118]

A solar panel are most efficient under natural sunlight, however, a solar panel can work using artificial light simply because a solar panel collects photons which collide with silicon atoms transferring their energy which cause them to lose electrons.

3 0

3 years ago

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