A., 101.7 km/h is the correct answer for this question
Free fall means your falling because of gravity and weight and will eventually hit the ground without any protection. weightlessness means a place or time where there is no gravity and you're floating and not falling.
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<u>Required</u><u> </u><u>Answer</u><u>:</u></h2>
The body will <u>stay at rest </u>(Option D). It is because a force of magnitude 50 N is pulled towards left and another force is pulling it towards right with same magnitude 50 N. So, the direction of force is opposite and magnitude is same i.e. 50 N. So, they will cancel each other and net force is 0. Hence, there would be no acceleration.
- Option A - Showing acceleration
- Option B - Showing acceleration
- Option C - Change of direction due to Net force
Hence, these options are incorrect because they are only possible when net external force is non-zero. Staying at rest i.e. Option D means there is no motion and hence no acceleration, this shows that net force is 0
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The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
Alternative solution:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Therefore, they have to fly 82.02 mph
Answer:
See description
Explanation:
With the given information we have:

the interval is ![[0,\pi ]](https://tex.z-dn.net/?f=%5B0%2C%5Cpi%20%5D)
now the mass
has the given expression:

we will use the formula for a line integral and let:

therefore we have:

we solve the integral:
