Answer:
hello the diagram relating to this question is attached below
a) angular accelerations : B1 = 180 rad/sec, B2 = 1080 rad/sec
b) Force exerted on B2 at P = 39.2 N
Explanation:
Given data:
Co = 150 N-m ,
<u>a) Determine the angular accelerations of B1 and B2 when couple is applied</u>
at point P ; Co = I* ∝B2'
150 = ( (2*0.5^2) / 3 ) * ∝B2
∴ ∝B2' = 900 rad/sec
hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec
at point 0 ; Co = Inet * ∝B1
150 = [ (2*0.5^2) / 3 + (2*0.5^2) / 3 + (2*0.5^2) ] * ∝B1
∴ ∝B1 = 180 rad/sec
hence angular acceleration of B1 = 180 rad/sec
<u>b) Determine the force exerted on B2 at P</u>
T2 = mB1g + T1 -------- ( 1 )
where ; T1 = mB2g ( at point p )
= 2 * 9.81 = 19.6 N
back to equation 1
T2 = (2 * 9.8 ) + 19.6 = 39.2 N
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Answer: the wavelength of this transition is 1.2039 um
Explanation:
Given that;
the energy level between the transitioning energy gap Eg = 1.0 + 0.03 = 1.03 eV
we know that λ = 1.24 / Eg
so we substitute our Eg into the above equation
λ = 1.24 / 1.03
λ = 1.2039 um
therefore the wavelength of this transition is 1.2039 um
Answer:
True
Explanation:
when the object is larger , the inertia of the object is larger so its tendency to change its state of motion is reduced is reduced.
Answer:
13.33 or 13 1/3m/s (meters per second)
Explanation:
In physics, we use the basic units of meters and seconds. So first convert (km) into meters (m) and also hours and minutes into seconds (s). We end up with 120000m and 9000s. Then divide the 120000m by the 9000s and you end up with 13.33 or 13 1/3 m/s.
After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is
Square root of ( 19.6 M ) .
If M=111 meters, then the speed is <em>46.64 meters per second</em>.
We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.
