<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2.
So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density.
So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave.
Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>
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>
Answer and Explanation:
A. We have temperature t = 32
Speed of sound, s = 1087.5
As t increases by 1⁰f speed increases by 1.2
So that
S = 1088.6
T= 33⁰f
We have 2 equations
1087.5 = k(32) + c
1088.6 = k(33) + c
Subtracting both equations
(33-32)k = 1088.6-1087.5
K = 1.1
b.). S = kT + c
1087.5 = 32(1.1) + c
Such that
C = 1052.3
Therefore
S = 1.1(t) + 1052.3
C.). S = 1.1t + 1052.3
We make t subject of the formula
T = s/1.1 - 1052.3/1.1
T = 0.90(s) - 956.3
D. This means that We have temperature to rise by 0.90 whenever speed is increased
Answer:
v = 66.4 m/s
Explanation:
As we know that plane is moving initially at speed of
now we have
now in Y direction we can use kinematics
since there is no acceleration in x direction so here in x direction velocity remains the same
so we will have
Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
