Answer:
14 m/s
Explanation:
Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.
Kinetic energy is given by ½mv²
Potential energy is given by mgh
Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.
Equating kinetic energy to be equal to potential energy then
½mv²=mgh
V
Making v the subject of the formula
v=√(2gh)
Substituting 9.81 m/s² for g and 10 m for h then
v=√(2*9.81*10)=14.0071410359145 m/s
Rounding off, v is approximately 14 m/s
To solve this problem it is necessary to apply the concepts related to Ohm's Law to find the voltage in the case of direct current. While for the Alternate current we use the concept of RMS voltage.
PART A) In the case of DC we have to,
V=IR
Where,
I = Current
R = Resistance
Where the current is 60mA y the resistance is
, then the potential would be


In the case of DC we have that the current and the RMS voltage is,



The potential in AC then is,



1. The problem statement, all variables and given/known data A person jumps from the roof of a house 3.4 meters high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 meters. If the mass of his torso (excluding legs) is 41 kg. A. Find his velocity just before his feet strike the ground. B. Find the average force exerted on his torso by his legs during deceleration. 2. Relevant equations I can't even seem to figure that part out. Help please? 3. The attempt at a solution I don't know how to start this at all
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω