Answer:
presumptuous
Explanation:
it's what you call someone who assumes something
Answer: P = 36.75W
The additional power needed to account for the loss is 36.75W.
Explanation:
Given;
Mass of the runner m= 60 kg
Height of the centre of gravity h= 0.5m
Acceleration due to gravity g= 9.8m/s
The potential energy of the body for each step is;
P.E = mgh
P.E = 60 × 9.8 × 0.5
PE = 294J
Since the average loss per compression on the leg is 10%.
Energy loss = 10% (P.E)
E = 10% of 294J
E = 29.4J
To calculate the runner's additional power
given that time per stride is = 0.8s
Power P = Energy/time
P = E/t
P = 29.4J/0.8s
P = 36.75W
ANSWER
Velocity of the mass reaches zero
EXPLANATION
We want to identify what hapens to a mass attached toa a spring at maximum displacement.
When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.
Hence, at maximum displacement, the velocity of the mass reaches zero.
Answer:
0.021 V
Explanation:
The average induced emf (E) can be calculated usgin the Faraday's Law:
<u>Where:</u>
<em>N = is the number of turns = 1 </em>
<em>ΔΦ = ΔB*A </em>
<em>Δt = is the time = 0.3 s </em>
<em>A = is the loop of wire area = πr² = πd²/4 </em>
<em>ΔB: is the magnetic field = (0 - 1.04) T </em>
Hence the average induced emf is:
Therefore, the average induced emf is 0.021 V.
I hope it helps you!
Answer:
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
Explanation:
Electrolytes are substances that, when are dissolved in solution, separates into electrical positive charges (cations) and electrical negative charges (anions) which are known as ions.
These ions have an adequate capacity to conduct particle charges and, therefore electricity.
Sodium, calcium, phosphate and potassium, are examples of electrolytes.
<u>Hence, the correct answer is:</u>
Electrolytes are considered ions when placed in a solution and allow for adequate conduction of particle charges.
I hope it helps you!