Can any juniors who go to Texas connections academy help me out with physics?

The acceleration of gravity (on Earth) is 9.8 m/s² downward.

This means that every falling object gains 9.8 m/s more downward speed
every second that it falls.

In 5 seconds of falling, it gains (5 x 9.8 m/s) = 49 m/s of downward speed.

If it was already descending at 2.0 m/s at the beginning of the 5 sec,
then at the end of the 5 sec it would be descending at

(2 m/s + 49 m/s) = 51 m/s .

7 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