Answer:
forward force of 4
Explanation:
Mathematically, this would be:
forward force of 6 - backward force of 2 = forward force of 4
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. Punctually we will verify the vertical displacement and the horizontal displacement from their respective components. We will start by calculating the time it took to reach the objective and later with that time, we will find the horizontal velocity launch component. The position can be written as,
![h= v_{0y}t+\frac{1}{2}a_yt^2](https://tex.z-dn.net/?f=h%3D%20v_%7B0y%7Dt%2B%5Cfrac%7B1%7D%7B2%7Da_yt%5E2)
Here,
h = Height
= Initial velocity in vertical direction
= Vertical acceleration (At this case, due to gravity)
= Time
There is not vertical velocity because the ball was thrown horizontally), then we have that
![6= (0)t+\frac{1}{2}(9.8)t^2](https://tex.z-dn.net/?f=6%3D%20%280%29t%2B%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2)
![t = 1.1065s](https://tex.z-dn.net/?f=t%20%3D%201.1065s)
Now using the equation of horizontal motion we have with this time that the initial velocity was,
![x = v_{ox}t+\frac{1}{2}a_xt^2](https://tex.z-dn.net/?f=x%20%3D%20%20v_%7Box%7Dt%2B%5Cfrac%7B1%7D%7B2%7Da_xt%5E2)
Here,
= Horizontal initial velocity
= Time
= Acceleration in horizontal plane
There is not acceleration in horizontal plane, only in vertical plane, then we have
![25= v_{0x}(1.1065)+\frac{1}{2}(0)(1.1065)^2](https://tex.z-dn.net/?f=25%3D%20v_%7B0x%7D%281.1065%29%2B%5Cfrac%7B1%7D%7B2%7D%280%29%281.1065%29%5E2)
![v_{0x} = 22.5938m/s](https://tex.z-dn.net/?f=v_%7B0x%7D%20%3D%2022.5938m%2Fs)
Therefore the pitching speed is 22.5938m/s
Answer:
Width = 680 [m]
Explanation:
To solve this problem we must know the speed of sound in the air, conducting an internet search, we find that this speed has a value of v = 340 [m/s].
We apply the following kinematic equation, which relates space to time, to define speed.
v = x /t
where:
x = distance [m]
t = time [s]
v = velocity [m/s]
Now replacing:
x = 340 * 4
x = 1360 [m]
But the width of the cannon is calculated when the sound wave hits the wall of the canyon, so this width is half the distance.
Width = 1360 / 2 = 680 [m]
Answer:
Yes, it's correct
Explanation:
Newton's second Law states that the acceleration of an object is proportional to the net force applied on it, according to the equation:
![F=ma](https://tex.z-dn.net/?f=F%3Dma)
where
F is the net force on the object
m is the mass of the object
a is the acceleration of the object
We can re-arrange the previous equation in order to solve explicitely for a, the acceleration, and we find:
![F=ma\\\frac{F}{m}=\frac{ma}{m}\\\frac{F}{m}=a\\a=\frac{F}{m}](https://tex.z-dn.net/?f=F%3Dma%5C%5C%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7Bma%7D%7Bm%7D%5C%5C%5Cfrac%7BF%7D%7Bm%7D%3Da%5C%5Ca%3D%5Cfrac%7BF%7D%7Bm%7D)
So, we see that the acceleration is proportional to the net force and inversely proportional to the mass of the object.
The uncertainty in a measured value is the possible variation in a measured value arising due to inaccuracies in the instrument being used. In the case described, the beats are 127 per minute, and the uncertainty is +/-4 beats. This means that the beats may actually be between 123 and 131.
The fractional uncertainty is: 4/127 = 0.031
And the percentage uncertainty is: 4/127 * 100 = 3.1%