Walking to school. Hope this helps!
The electric field at (x,y,z) its equal to the negative of the gradient if the electric potential.
We have
V(x, y, z) = 2x² - 3y² + 5z
so
E(x, y, z) = -grad(V) = -(dV/dx i + dV/dy j + dV/dz k)
where d/d(variable) is meant to be a partial derivative with respect to that variable. The partial derivatives are
dV/dx = 4x
dV/dy = -6y
dV/dz = 5
and so the electric field at any point is
E(x, y, z) = -4x i + 6y j - 5k
and at (3, 4, 5) it is
E(3, 4, 5) = -12i + 24j - 5k
Answer:
The shearing stress is 10208.3333 Pa
The shearing strain is 0.25
The shear modulus is 40833.3332 Pa
Explanation:
Given:
Block of gelatin of 120 mm x 120 mm by 40 mm
F = force = 49 N
Displacement = 10 mm
Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?
The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:
A = 40*120 = 4800 mm² = 0.0048 m²
The shearing stress:

The shearing strain is defined as the tangent of the displacement that the block over its length:

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f =
D_f =
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
Answer:Reducing mass i.e. water
Explanation:
Frequency For given mass in glass is given by

where k =stiffness of the glass
m=mass of water in glass
from the above expression we can see that if mass is inversely Proportional to frequency
thus reducing mass we can increase frequency