Answer:
Explanation:
The balloon would require a time of
t = d/v = 13.5/ (23.6cos38) = 0.7259...s
to travel the horizontal distance.
the vertical position relative to the throw point at that time is
h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)
h = 7.9652...
so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.
If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time
Answer:
28.3 kg
Explanation:
Assuming the ground is level, the normal force equals the weight.
N = mg
277 N = m × 9.8 m/s²
m = 28.3 kg
We assume that the gas is an ideal gas so we can use the relation PV=nRT. Assuming that the temperature of the system is at ambient temperature, T = 298 K. We can calculate as follows:
PV = nRT
P = nRT / V
P = (0.801 mol ) (0.08205 L-atm / mol-K) (298.15 K) / 12 L
P = 1.633 atm
= t(Vf + Vi/2)
<span>Vf + Vi/2 = d/t </span>
<span>Vf = (d/t) - Vi/2 </span>
<span>Answer: Vf = (d/t) - Vi/2 OR (2dt - Vit)/2t
</span>
Answer:
4.09 kgm/s
Explanation:
Concept tested: Second Newton's law of motion
So what does the second Newton's law of motion state?
- According to the second Newton's law of motion, the resultant force and the rate of change of momentum are directly proportional.
- That is;
, where, mVf is the final momentum and mVo is the initial momentum.
In this case;
- Force = 12.8 N
- Initial momentum, mVo = 1.3 kgm/s
- Time, t = 0.218 seconds
Therefore; replacing the values of the variables in the formula;
we get;
Solving for final momentum;
= 4.09 kgm/s
Therefore, the final momentum of the box is 4.09 kgm/s
