<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B
<em>I hope this helps you </em>
Answer:
1) Newton's first law of motion states an object will remain at rest or in uniform will be in uniform motion in a straight line unless a force acts on it
2) Newton's second law states the acceleration of an object is directly proportional to the applied force acting on an object and inversely proportional to the mass of the object
Explanation:
1) With Newton's first law, we are able arrange things within a space and schedule meetings in time knowing that they will remain in place unless an external force changes their positions
2) An example of Newton's second law of motion is that small objects such as a ball are easily accelerated and can be given appreciable acceleration for flight by single, one time contact (such as kicking the ball) while larger objects such as a rock require sustained force application to change their location.
Are there any options or is it not multiple choice.
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
<span>2002 seconds, or 33 minutes, 22 seconds.
First, let's calculate how many joules it will take to lift 78 kg against gravity for 1100 meters. So:
78 kg * 9.8 m/s^2 * 1100 m = 840840 kg*m^2/s^2
Now a watt is defined as kg*m^2/s^3, so a division of the required joules should give us a convenient value of seconds. So:
840840 kg*m^2/s^2 / 420 kg*m^2/s^3 = 2002 seconds.
And 2002 seconds is the same as 33 minutes, 22 seconds.</span>
