Answer:
28.79%
Explanation:
Given
Design Speed, V = 120km/h = 33.33m/s
Radius, R = 300m
Side Friction, Fs = 0.09
Gravitational Constant = 9.8m/s²
Using the following formula, we'll solve the required rate of superelevation.
e + Fs = V²/gR where e = rate
e = V²/gR - Fs
e = (33.33)²/(9.8 * 300) - 0.09
e = 0.287853367346938
e = 28.79%
Hence, the required rate of superelevation for the curve is calculated as 28.79%
Answer:
Doubled
Explanation:
F = (kq1q2) / r^2
F and q (Either q1 or q2) are directly proportional, so double the charge would also double the electruc force between the charges.
Answer:
The work done by the gravel to stop the truck is 520.44 kJ
Explanation:
<u>Step 1</u>: Data given
Mass of the truck = 3047.8 kg
The ramp has an angle of 9.5 °
Velocity of the truck = 20.68 m/s
distance = 26.6 meters
<u>Step 2:</u> Calculate initial kinetic energy
sin 9.5° = 0.165
h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m
Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ = initial kinetic energy
<u>Step 3: </u>Calculate potential energy
Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ
<u>Step 4:</u> What work is done by the truck on the gravel?
Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ
Answer:
(a) 6650246.305 N/C
(b) 24150268.34 N/C
(c) 6408227.848 N/C
(d) 665024.6305 N/C
Explanation:
Given:
Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]
Total charge of the ring (Q) = 75.0 μC = [1 μC = 10⁻⁶ C]
Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:
Plug in the given values for each point and solve.
(a)
Given:
,
Electric field is given as:
(b)
Given:
,
Electric field is given as:
(c)
Given:
,
Electric field is given as:
(d)
Given:
,
Electric field is given as:
Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol
<h3>
<em>Al(OH)3 = 78.01 g/mol</em></h3>