A water wave that occurs in an ocean is an example of which type of wave?
1 answer:
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Answer:
54.9 m/s at 44.9 degrees
Explanation:
If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.
Vx = V * cos(a) = 46.2 * cos(-32.7) = 38.9 m/s
Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s
SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.
Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use
Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2
We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m
If we take the first derivative of the equation of position, we get the equation for speed
V(t) = Vy0 + a * t
We know that being t2 the moment the ball goes over the wall
V(t2) = -25 m/s
Y(t2) = 45.4 m
So:
45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2
-25 = Vy0 + a * t2
Then:
Vy0 = -25 - a * t2
So:
45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2
0 = -44.4 - 25 * t2 - 1/2 * a * t2^2
a = -9.81 m/s^2
0 = -44.4 - 25 * t2 + 4.9 * t2^2
Solving this quadratic equation we get:
t1 = -1.39 s
t2 = 6.5 s
Since we are looking for a positive value we disregard t1.
Now we can obtain Vy0:
Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s
Since horizontal speed is constant Vx0 = 38.9 m/s
By Pythagoras theorem we obtain the value of the initial speed:
The angle is in the the first quadrant because both comonents ate positive, so: 0 < a < 90
a = atan(Vy0/Vx0) = 44.9 degrees
Answer:
16.935 N
Explanation:
In order to make the box start moving, the level force applied on the box (F) must be greater than the force of static friction that keeps the box at rest, which is equal to
where
is the coefficient of static friction
(mg) = 30 N is the weight of the box
Therefore, the condition for F must be:
So, the applied force must be greater than this value.