Answer:
C = 4,174 10³ V / m^{3/4}
, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C 
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
<em>The convex lens is a lens that converges rays of light that convey parallel to its principal axis (i.e. converges the incident rays towards the principal axis) which is relatively thick across the middle and thin at the lower and upper edges. The edges are curved outward rather than inward.</em>
Answer: FR=2.330kN
Explanation:
Write down x and y components.
Fx= FSin30°
Fy= FCos30°
Choose the forces acting up and right as positive.
∑(FR) =∑(Fx )
(FR) x= 5-Fsin30°= 5-0.5F
(FR) y= Fcos30°-4= 0.8660-F
Use Pythagoras theorem
F2R= √F2-11.93F+41
Differentiate both sides
2FRdFR/dF= 2F- 11.93
Set dFR/dF to 0
2F= 11.93
F= 5.964kN
Substitute value back into FR
FR= √F2(F square) - 11.93F + 41
FR=√(5.964)(5.964)-11.93(5.964)+41
FR= 2.330kN
The minimum force is 2.330kN
Answer:
the minimum value of torque is 101.7 N-m.
Explanation:
Given that,
Mass of bundle of shingles, m = 30 kg
Upward acceleration of the shingles,
The radius of the motor of the pulley, r = 0.3 m
Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :
T-mg=ma
T=m(g+a)
Let 
is the minimum torque that the motor must be able to provide. It is given by :
the minimum value of torque is 
Answer:
t = 13.43 s
Explanation:
In order to find the minimum time required by the plane to stop, we will use the first equation of motion. The first equation of motion is written as follows:
Vf = Vi + at
where,
Vf = Final Velocity of the Plane = 0 m/s (Since, the plane finally stops)
Vi = Initial Velocity of the Plane = 95 m/s
a = deceleration of the plane = - 7.07 m/s²
t = minimum time interval needed to stop the plane = ?
Therefore,
0 m/s = 95 m/s + (- 7.07 m/s²)t
t = (95 m/s)/(7.07 m/s²)
<u>t = 13.43 s</u>
