Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:
Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)
We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2
b) The acceleration of point A is:
(aA/D)t = ADαj
The acceleration of point E is:
(aE/D)t = -EDαj
Answer:
Explanation:
Magnitude of force per unit length of wire on each of wires
= μ₀ x 2 i₁ x i₂ / 4π r where i₁ and i₂ are current in the two wires , r is distance between the two and μ₀ is permeability .
Putting the values ,
force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )
= .67 i² x 10⁻⁴
force on 3 m length
= 3 x .67 x 10⁻⁴ i²
Given ,
8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²
i² = 3.98 x 10⁻²
i = 1.995 x 10⁻¹
= .1995
= 0.2 A approx .
2 i = .4 A Ans .
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals 
meters
You don't constantly feel your clothes on you because of sensory adaption. So basically your body got used to it.
(I'm lucky to have a computer ... It was only through the miracle of
modern digital technology that I was able to flip your photo right-
side-up to where I could read it.)
Here's how to figure out things like this:
The circle on the left side labeled ' <em>G</em> ' is the <em><u>G</u></em>enerator or battery
that powers this whole circuit and all the devices in it. In order for
any device to work, you need to be able to set your pencil down at
the top of the Generator, and find a path through the circuit and
through that device, where current can flow all the way around to
the bottom of the Generator. If you ever come to an open switch,
then current stops there, and you have to find another way through.
If the path you found takes you back to the bottom of the generator but
it doesn't go through one of the devices, then that device doesn't work.
Look at the picture. If you open switch S-4, then Device-4 can't work,
because current can't go through it from one end of the Generator to
the other end. But all of the other devices still work.
I can see 2 ways to turn off Device-3 with a single switch ... either
open switch S-5, or else open switch S-1. Unfortunately, I think
either way will shut off all 5 devices.
