Answer:
The value is
Explanation:
From the we are told that
The initial speed of the object is
The greatest height it reached is 
Generally from kinematic equation we have that

At maximum height v = 0 m/s
So

=> 
Here H is the height from the initial height to the maximum height
So the initial height is mathematically represented as

=> 
=> 
Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

At maximum height v = 0 m/s

=> 
Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

Here the initial velocity is 0 m/s given that its the velocity at maximum height
Also g is positive because we are moving in the direction of gravity
So

=> 
Generally the total time taken is mathematically represented as

=> 
=>
Answer:
The answer is false
Explanation:
Though the mostly used SI unit of measurement or the most popular units are the
Length,
Time and
Mass
i.e meter (m), seconds (s), kilogram (kg)
Aside all the above stated units for measurements there are other four basic units which are itemized bellow.
they are
1. Amount of substance - mole (mole)
2. Electric current - ampere (A)
3. Temperature - kelvin (K)
4. Luminous intensity - candela (cd)
Answer:
The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Explanation:
Given:
(A)
Current
A
Voltage
V
For finding the resistance,



12Ω
(B)
For finding power delivered,


Watt
(C)
For finding the potential difference,



V
Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J