Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.
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Answer:
Explanation:
Given
radius of circular path
Position is given by
Differentiate 1 to angular velocity we get
Differentiate 2 to get angular acceleration
Net acceleration is the vector summation of tangential and centripetal force
A procedure is all the steps used to do an experiment in order.
<span>the experiment is when you test your hypothesis and is designed to answer your question. </span>
<span>the procedure is all the steps of the experiment.</span>
Answer:
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g
U is the speed = 300m/s
H is the maximum height = 78.4m
g is the acceleration due to gravity = 9.8m/s²
Substitute into the fromula;
R = 300√2(78.4)/9.8
R = 300 √(16)
R = 300*4
R = 1200m
Hence the projectile travelled 1200m before hitting the ground
Answer:
Av = 25 [m/s]
Explanation:
To solve this problem we must use the definition of speed, which is defined as the relationship between distance over time. for this case we have.

where:
Av = speed [km/h] or [m/s]
distance = 180 [km]
time = 2 [hr]
Therefore the speed is equal to:
![Av = \frac{180}{2} \\Av = 90 [km/h]](https://tex.z-dn.net/?f=Av%20%3D%20%5Cfrac%7B180%7D%7B2%7D%20%5C%5CAv%20%3D%2090%20%5Bkm%2Fh%5D)
Now we must convert from kilometers per hour to meters per second
![90[\frac{km}{h}]*1000[\frac{m}{1km}]*1[\frac{h}{3600s} ]= 25 [m/s]](https://tex.z-dn.net/?f=90%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A1000%5B%5Cfrac%7Bm%7D%7B1km%7D%5D%2A1%5B%5Cfrac%7Bh%7D%7B3600s%7D%20%5D%3D%2025%20%5Bm%2Fs%5D)