At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
1 answer:
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Average velocity = Total Displacement / Total time taken
Total displacement is the distance from starting to end point.
35km west then 90km north. When you will draw this and connect starting to end point. You will get a right angled triangle.
So displacement =
= 96.56 km
So, Total displacement = 96.56 km
So, Average Velocity = 96.56 / 2 = 48.28 km/hr
Total displacement is the distance from starting to end point.
35km west then 90km north. When you will draw this and connect starting to end point. You will get a right angled triangle.
So displacement =
= 96.56 km
So, Total displacement = 96.56 km
So, Average Velocity = 96.56 / 2 = 48.28 km/hr
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα=
Senα=
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s
from the question you can see that some detail is missing, using search engines i was able to get a similar question on "https://www.slader.com/discussion/question/a-student-throws-a-water-balloon-vertically-downward-from-the-top-of-a-building-the-balloon-leaves-t/"
here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity
v = 15.4 m/s