If and
are perpendicular, then their dot product is zero. This means
Solving for n is trivial; it follows that n = 3.
PLEASE I NEED SOME HELP!
1 answer:
You might be interested in
The force of friction is equal to the pushing force, and the acceleration is zero
Explanation:
In this problem, we are pushing a piano along the floor, in the horizontal direction.
There are 2 forces acting in the horizontal direction on the piano:
- The applied force,
, acting forward
- The force of friction,
, acting backward
Therefore, the net force in the horizontal direction is
According to Newton's second law, the net force is equal to the product between the piano's mass (m) and its acceleration:
Combining the two equations,
However, we are also told that the piano moves at constant speed, therefore the acceleration is zero:
And so,
which means that the force of friction is equal to the applied force.
Learn more about Newton's second law:
#LearnwithBrainly
Refer to the diagram shown below.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.