The correct answer for the volume v of the solid obtained by rotating the region bounded by the given curves about the specified line. y = x2, y = 4x is 128/3 * π.
Volume of Solid of Revolution by Shell method is given by
V = 2π * integrate x(height) dx Here, height = 4x-x2
(1)& x-varies from x = 0 to x = 4 then from eqn(1) V = 2π * integrate x(4x - x ^ 2) dx from x = 0 to 4 = 2π * integrate (4x ^ 2 - x ^ 3) dx from x = 0 to 4
Basic Rule(1) ∫ x^n dx =x^ n+1/ n+1
V=2 π [4((x ^ 3)/3) - (x ^ 4)/4] 0 ^ 4 =2 π[ 4/3 x^ 3 - x^ 4/4 ] 0 ^ 4
V = 2π [4/3 * 4 ^ 3 - (4 ^ 4)/4} - 0]
V = 128/3 * π.
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Kinetic energy= .5 x m x v^2
KE=.5 x 4.2 x 3.85^2
KE=31.13
Answer:
14m/s
Explanation:
Given parameters:
Radius of the curve = 50m
Centripetal acceleration = 3.92m/s²
Unknown:
Speed needed to keep the car on the curve = ?
Solution:
The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.
It is given as;
a =
a is the centripetal acceleration
v is the speed
r is the radius
Now insert the parameters and find v;
v² = ar
v² = 3.92 x 50 = 196
v = √196 = 14m/s
Answer:
t = 6.09 seconds
Explanation:
Given that,
Speed, v = 44.1 cm/s
Distance, d = 269 cm
We need to find the time interval of the marble. Speed is distance per unit time.

Hence, the time interval of the marble is 6.09 seconds.
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;

E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C