You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 3.5 H
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Answer:
5.7 m
Explanation:
AD = length of the ladder = L = 8 m
AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m
AC = distance of person on the ladder from the bottom end = x
W = weight of the ladder = 240 N
= weight of the person = 710 N
F = force by the wall on the ladder
N = normal force by ground on the ladder = ?
Using equilibrium of force along the vertical direction
N = + W
N = 710 + 240
N = 950 N
μ = Coefficient of static friction = 0.55
f =static frictional force on the ladder
Static frictional force is given as
f = μ N
f = (0.55) (950)
f = 522.5 N
Force equation along the horizontal direction is given as
F = f
F = 522.5 N
using equilibrium of torque about point A
F Sin50 (AD) = W Cos50 (AB) + ( Cos50 (AC))
(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)
x = 5.7 m
Answer:
y = - (½ g / v₀²) x²
Explanation:
This is a projectile launch exercise where there is no acceleration on the x-axis so
x = v₀ₓ t
v₀ₓ = v₀ cos tea
y = t - ½ g t2
v_{oy} = v₀ sin θ
as the sphere is thrown horizontally, the angle is tea = 0º, so the initial velocity remains
v₀ₓ = v₀
v_{oy} = 0
we substitute in our equations
x = v₀ t
y = - ½ g t²
we eliminate the time from these equations, we substitute the first in the second
y = - ½ g (x / v₀)²
y = - (½ g / v₀²) x²
this is the equation of a parabola
Answer:
a) E = 8.63 10³ N /C, E = 7.49 10³ N/C
b) E= 0 N/C, E = 7.49 10³ N/C
Explanation:
a) For this exercise we can use Gauss's law
Ф = ∫ E. dA = /ε₀
We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product
The area of a sphere is
A = 4π r²
if we use the concept of density
ρ = q_{int} / V
q_{int} = ρ V
the volume of the sphere is
V = 4/3 π r³
we substitute
E 4π r² = ρ (4/3 π r³) /ε₀
E = ρ r / 3ε₀
the density is
ρ = Q / V
V = 4/3 π a³
E = Q 3 / (4π a³) r / 3ε₀
k = 1 / 4π ε₀
E = k Q r / a³
let's calculate
for r = 4.00cm = 0.04m
E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³
E = 8.63 10³ N / c
for r = 6.00 cm
in this case the gaussine surface is outside the sphere, so all the charge is inside
E (4π r²) = Q /ε₀
E = k q / r²
let's calculate
E = 8.99 10⁹ 3 10⁻⁹ / 0.06²
E = 7.49 10³ N/C
b) We repeat in calculation for a conducting sphere.
For r = 4 cm
In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.
E = 0
In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is
E = k q / r²
E = 7.49 10³ N / C