Answer:
Explanation:
a) series resistors carry the same current
A = V/Re = 6/(16 + 6) = 0.2727272... ≈ 27 mA
b) V = V₀(R/Re) = 6(16/(16 + 6)) = 4.363636 ≈ 4.4 V
c) V = V₀(R/Re) = 6(6/(16 + 6)) = 1.636363 ≈ 1.6 V
or V = 6 - 4.4 = 1.6 V
Answer:
Part a)

Part b)
v = 3.64 m/s
Part c)

Part d)

Explanation:
As we know that moment of inertia of hollow sphere is given as

here we know that

R = 0.200 m
now we have


now we know that total Kinetic energy is given as





Part a)
Now initial rotational kinetic energy is given as



Part b)
speed of the sphere is given as
v = 3.64 m/s
Part c)
By energy conservation of the rolling sphere we can say




Part d)
Now we know that




Answer:

Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )


An ampere (AM-pir), or amp
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