Answer:
25.0 m / 10 m/s = 2.5 s
50.0 / 9.5 = 5.26 s
25.0 / 11.1 = 2.25 s
T (Wood) = 2.5 + 5.26 + 2.25 = 10.0 s
Mrs Wood runs 10 s vs 10.4 for Mr Overstreet
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Objects in free fall, disregarding terminal velocity, accelerate at 9.8(m/s)/s. so for every second it was falling, it gained 9.8m/s in speed. 9.8 * 10 = 98m/s
The correct answer to the question is : Electric energy
EXPLANATION :
As per the question, we have an electric generator.
Before coming into any conclusion, first we have to understand the function of generator.
The generator is attached to the turbine. When the turbine rotates, the generator also starts rotating with it. Thanks to electromagnetic induction, the electricity is produced in the coil attached to the generator when it rotates.
Hence, from above, it is obvious that kinetic energy is converted into electric energy.
Answer:
K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where,
G is gravitational constant
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U
