Question

both mass and weight decrease
both mass and weight stay the same
5 0/1 point formula
a 63 kg student is riding in an elevator that is traveling upward but slowing down at it reaches the top floor. if the acceleration of the elevator while it slows has a magnitude of 2.7 m/s/s, what is the apparent weight of the student during this time? round your answer to 1 decimal place if necessary.
6 1/1 point formula
a car (mass 1183 kg) traveling with a speed of 37 m/s comes to a stop on a horizontal surface over a distance of 53 m. what is average force exerted on the car by the brakes during this time? round your answer to the nearest whole number.

Explanation:

Step1: Identify given values

$m = 63\ \text{kg}$, $a = 2.7\ \text{m/s}^2$ (downward, so negative for upward motion), $g = 9.8\ \text{m/s}^2$

Step2: Use apparent weight formula

Apparent weight $W_{app} = m(g - a)$

Step3: Substitute values

$W_{app} = 63\times(9.8 - 2.7)$

Step4: Calculate result

$W_{app} = 63\times7.1 = 447.3$
Wait, correction: Wait, $9.8 - 2.7 = 7.1$, $63*7.1 = 447.3$? No, wait $63*7 = 441$, $63*0.1=6.3$, total $441+6.3=447.3$? Wait no, wait when elevator slows upward, acceleration is downward, so net force is $mg - N = ma$, so $N = m(g - a)$. Yes, $g=9.8$, so $9.8-2.7=7.1$, $63*7.1=447.3$. Wait, maybe using $g=9.81$? Let's recalculate:
$9.81-2.7=7.11$, $63*7.11=63*(7+0.11)=441 + 6.93=447.93≈447.9$. Wait, but maybe I messed up sign. Wait no: apparent weight is the normal force. When elevator decelerates upward, acceleration is downward, so $F_{net} = mg - N = ma$, so $N = m(g - a)$. Yes. So if $g=9.8$, $63*(9.8-2.7)=63*7.1=447.3$. If $g=9.81$, $63*(9.81-2.7)=63*7.11=447.93≈447.9$. Wait, maybe the original wrong answer was 170.1 which was way too low. Oh wait, no! Wait, 63*2.7 is 170.1, that's the mistake: they used $ma$ instead of $m(g-a)$. So correct answer is 447.3 N (or 447.9 if using 9.81 for g). Let's use $g=9.8$ for standard:

Step1: List known values

$m=63\ \text{kg}$, $a=2.7\ \text{m/s}^2$ (downward), $g=9.8\ \text{m/s}^2$

Step2: Apparent weight formula

$W_{app}=m(g-a)$

Step3: Plug in numbers

$W_{app}=63\times(9.8-2.7)$

Step4: Compute

$W_{app}=63\times7.1=447.3$

Final answer: 447.3 N (or 447.9 N with g=9.81, but 447.3 is standard with g=9.8)

Answer:

454.2 N