Position is measured in meters (m), so it is a base quantity.
<h3>What is base quantity?</h3>
A base or fundamental quantity is a physical quantity, in which other quantities are derived from.
Example of fundamental quantities;
- Mass
- Length (position)
- Time
- Temperature
- Amount of substance
<h3>What is a derived quantity?</h3>
Derived quantities are those quantities obtained or expressed from fundamental quantities.
Example of derived quantities;
- Speed
- Acceleration
- Volume
- Area
- Density, etc
Thus, we can conclude that position measured in meters (m) is a base quantity.
Learn more about base quantities here: brainly.com/question/14480063
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Answer:
23. 4375 m
Explanation:
There are two parts of the rocket's motion
1 ) accelerating (assume it goes upto h1 height )
using motion equations upwards

Lets find the velocity after 2.5 seconds (V1)
V = U +at
V1 = 0 +5*2.5 = 12.5 m/s
2) motion under gravity (assume it goes upto h2 height )
now there no acceleration from the rocket. it is now subjected to the gravity
using motion equations upwards (assuming g= 10m/s² downwards)
V²= U² +2as
0 = 12.5²+2*(-10)*h2
h2 = 7.8125 m
maximum height = h1 + h2
= 15.625 + 7.8125
= 23. 4375 m
Answer:
T = 764.41 N
Explanation:
In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:
(1)
m: mass object = 2.3 kg
r: radius of the circular orbit = 0.034 m
v: tangential speed of the object
However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

you replace this value of v in the equation (1). Also, you replace the values of r and m:

hence, the tension in the string must be T = Fc = 764.41 N
Answer:
The fundamental frequency of can is 2.7 kHz.
Explanation:
Given that,
A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm
The speed of sound is, v = 336 m/s
We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.