<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>
Answer:
D40 = 2.56 × D25
so number is 2.56 multiple of stopping distance @ 25 mph
Explanation:
given data
speed = 40 miles / hour
distance = D40
speed limit = 25 miles / hour
distance = D25
to find out
express number a multiple of stopping distance @ 25 mph
solution
we know here stopping distance is directly proportional to (speed)²
so here speed ratio is
initial speed =
so initial speed = 1.6
so
stopping distance increase = (1.6)²
= (1.6)²
= 2.56
so here
D40 = 2.56 × D25
so number is 2.56 multiple of stopping distance @ 25 mph
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is
Explanation:
From the question we are told that
The third charge is 
The position of the third charge is 
The first charge is 
The position of the first charge is 
The second charge is 
The position of the second charge is
The distance between the first and the third charge is


The force exerted on the third charge by the first is

Where k is the coulomb's constant with a value 
substituting values
The distance between the second and the third charge is


The force exerted on the third charge by the first is mathematically evaluated as
substituting values

The net force is
substituting values

Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
By v^2 = u^2 + 2gh
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s