A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
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Answer:
Explanation:
If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.
The distance between the 2 crests (or troughs) is the wavelength, λ.
We want the velocity of this wave. The equation that relates these 3 things is
and filling in:
so
v = 2.5(2.0) and
v = 5.0 m/s
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