Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
<em><u>Support Cy:</u></em>
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
<em><u>Support Ay:</u></em>
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
Explanation:
First of all we shall find the velocity at equilibrium point of mass 1.2 kg .
It will be ω A , where ω is angular frequency and A is amplitude .
ω = √ ( k / m )
= √ (170 / 1.2 )
= 11.90 rad /s
amplitude A = .045 m
velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s
= .5355 m /s
At middle point , no force acts so we can apply law of conservation of momentum
m₁ v₁ = ( m₁ + m₂ ) v
1.2 x .5355 = ( 1.2 + .48 ) x v
v = .3825 m /s
= 38.25 cm /s
Let new amplitude be A₁ .
1/2 m v² = 1/2 k A₁²
( 1.2 + .48 ) x v² = 170 x A₁²
( 1.2 + .48 ) x .3825² = 170 x A₁²
A₁ = .0379 m
New amplitude is .0379 m
The line at the bottom of the picture ... probably the first line on a list of choices .. is the correct equation.
Answer:
nuclear battery to generate energy
