What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
Answer:
a) F = 64.30 N, b) θ = 121.4º
Explanation:
Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components
let's use trigonometry
Force F1
sin 170 = F_{1y} / F₁
cos 170 = F₁ₓ / F₁
F_{1y} = F₁ sin 170
F₁ₓ = F₁ cos 170
F_{1y} = 100 sin 170 = 17.36 N
F₁ₓ = 100 cos 170 = -98.48 N
Force F2
sin 30 = F_{2y} / F₂
cos 30 = F₂ₓ / F₂
F_{2y} = F₂ sin 30
F₂ₓ = F₂ cos 30
F_{2y} = 75 sin 30 = 37.5 N
F₂ₓ = 75 cos 30 = 64.95 N
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = -98.48 +64.95
Fₓ = -33.53 N
Y axis
F_y = F_{1y} + F_{2y}
F_y = 17.36 + 37.5
F_y = 54.86 N
a) the magnitude of the resultant vector
let's use Pythagoras' theorem
F = Ra Fx ^ 2 + Fy²
F = Ra 33.53² + 54.86²
F = 64.30 N
b) the direction of the resultant
let's use trigonometry
tan θ’= F_y / Fₓ
θ'= 
θ'= tan⁻¹ (54.86 / (33.53)
θ’= 58.6º
this angle is in the second quadrant
The angle measured from the positive side of the x-axis is
θ = 180 -θ'
θ = 180- 58.6
θ = 121.4º
Answer:
charge on each
Q1 = 2.06 ×
C
Q2 = 7.23 ×
C
when force were attractive
Q1 = 1.07 ×
C
Q2 = -1.39 ×
C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 ×
C
and
According to Coulomb's law force between two sphere is
Force F =
.........1
Q1Q2 = 
here F is force and r is apart distance and k is 9 ×
N-m²/C² put all value we get
Q1Q2 = 
Q1Q2 = 1.49 ×
C²
and
we have Q2 = 93 ×
C - Q1
put here value
Q1² - 93 ×
Q1 + 1.49 ×
= 0
solve we get
Q1 = 2.06 ×
C
and
Q1Q2 = 1.49 ×
2.06 ×
Q2 = 1.49 ×
Q2 = 7.23 ×
C
and
if force is attractive we get here
Q1Q2 = - 1.49 ×
C²
then
Q1² - 93 ×
Q1 - 1.49 ×
= 0
we get here
Q1 = 1.07 ×
C
and
Q1Q2 = - 1.49 ×
2.06 ×
Q2 = - 1.49 × 
Q2 = -1.39 ×
C
Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1