To solve this problem we must use the trigonometric relations, in this case as we have the ramp and the height the concentric trigonometric property to the will allow us to find the angle.
The sine of angle A, is defined as
Therefore the angle of elevation is 8.9°
Answer:
The new velocity of the string is 100 centimeters per second (1 meter per second).
Explanation:
The speed of a wave through a string (), in meters per second, is defined by the following formula:
(1)
Where:
- Tension, in newtons.
- Length of the string, in meters.
- Mass of the string, in kilograms.
The expression for initial and final speeds of the wave are:
Initial speed
(2)
Final speed
(3)
By (2), we conclude that:
If we know that , then the new speed of the wave in the string is
.
Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3 = 5 3 2
x_{f} = 45 m
we substitute
v = 15 m / s
Answer:
245.45km in a direction 21.45° west of north from city A
Explanation:
Let's place the origin of a coordinate system at city A.
The final position of the airplane is given by:
rf = ra + rb + rc where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:
rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km
rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km
The module of this position is:
And the angle measure from the y-axis is:
So the answer is 245.45km in a direction 21.45° west of north from city A