How often would you push someone on a swing to create a frequency of .4 hertz
1 answer:
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Answer:
Acceleration,
Explanation:
Initial velocity of a particle in vector form, u = (-5i - 2j) m/s
Final velocity of particle in vector form, v = (-6i + 7j) m/s
Time taken, t = 8 seconds
We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :
or
Hence, the value of acceleration vector is solved.
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity:
where
is the final velocity
is the initial velocity
is the acceleration due to gravity 
is the time
2. Kinematic equation for distance:
where
is the distance
is the initial velocity
is the final velocity
is the acceleration due to gravity
is the time
First, we are going to convert 21.82 mi/h to ft/s:
Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore:
. We also know that it initial speed is 31.21 ft/s, so 
. Lets replace those values in our formula to find :
seconds
Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
ft
Now we can add the height of the building and the maximum height of the object:
ft
Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
or 
Since time cannot be negative,
is the time it takes the object to fall to the ground.
Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
ft/s
We can conclude that the speed of the object when it hits the ground is 110.25 ft/s
1. Kinematic equation for final velocity:
where
2. Kinematic equation for distance:
where
First, we are going to convert 21.82 mi/h to ft/s:
Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore:
Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
Now we can add the height of the building and the maximum height of the object:
Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
Since time cannot be negative,
Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
We can conclude that the speed of the object when it hits the ground is 110.25 ft/s
ANSWER
EXPLANATION
Parameters given:
Mass of the student, M = 70 kg
Mass of the textbook, m = 1 kg
Distance, r = 1 m
To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:
where G = gravitational constant
Therefore, the gravitational force acting between the student and the textbook is:
That is the answer.