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Lina20 [59]
3 years ago
14

What two types of scientific knowledge can be expressed as mathematical equations? (Please don't tell me the answer could you pl

ease just follow me there. Thank you!)
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0
An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.
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Shadow and eclipses result from​
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If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{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: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
0=31.21+(-32)t
-32t=-31.21
t= \frac{-31.21}{-32}
t=0.98seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
d=15.22ft 

Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

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:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
16t^2+31.21t-175.22=0
t=2.47 or t=-4.43
Since time cannot be negative, t=2.47 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:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


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What are facts about balloons?
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Answer:

I think your soppose to multiply

Explanation:

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