5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5
<h3>What is
displacement ?</h3>
A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.
The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector
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1- You should always have a question for your experiment.
2- You need to conduct background research. It helps to write down your sources so you can cite your references.
3- Propose a hypothesis (educated guess on what you believe the outcome of the experiment will be)
4- Design and perform an experiment to test your hypothesis (include independent and dependent variable)
5- Record observations and analyze what the data means.
6- Conclude whether you need to accept or reject your hypothesis, which accepting means your hypothesis was right and rejected is if it was wrong.
The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
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Answer:
56.7°
Explanation:
Imagine a rectangle triangle.
The triangle has 3 sides.
One side is the height of the tower, let's name it A.
Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.
Sides A and B are perpendicular.
The other side is the length of the wire. Let's name it C.
From trigonometry we know that:
cos(a) = B / C
Where a is the angle between B anc C, between the wire and the ground.
Therefore
a = arccos(B/C)
a = arccos(552/1005) = 56.7°
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
