<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth
Explanation:
The period of a simple pendulum is given by the equation
where
L is the length of the pendulum
g is the acceleration of gravity
In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is
where
is the acceleration of gravity on Earth
The period of the pendulum on the Moon is
where
is the acceleration of gravity on the Moon
Calculating the ratio of the period on the Moon to the period on the Earth, we find
Therefore, for every hour interval on Earth, the Moon clock will display a time of
A) (9.8/1.6)h
#LearnwithBrainly
· free fall is any motion of a body where gravity is the only forceacting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it.
<span>15 m/s^2
The first thing to calculate is the difference between the final and initial velocities. So
180 m/s - 120 m/s = 60 m/s
So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving
60 m/s / 4.0 s = 15.0 m/s^2
Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2</span>
False. That description fits the wave's 'frequency'.
It has nothing to do with refraction.
