Answer:
5.3 × 10^(-8) m
Explanation:
We are given;
Shear force; F = 400 N
Length of cube; L_o = 30 cm = 0.3 m
Shear modulus; S = 2.5 × 10^(10) N/m²
Now,the resulting relative displacement can be gotten from the formula;
F = A × S × Δx/L_o
Where Δx is resulting relative displacement
A is area.
Area of cube = (L_o)² = 0.3² = 0.09
Thus, making Δx the subject, we have;
Δx = (F × L_o)/(A × S)
Plugging in the relevant values;
Δx = (400 × 0.3)/(0.09 × 2.5 × 10^(10))
Δx = 5.3 × 10^(-8) m
<span>conservation of momentum
mass(ball) × speed(ball) = mass(cannon) × speed(cannon)
(8.3 kg) × (187.9 m/s) = (949.3 kg) × v
1.64 m/s = v
</span>
Answer:
<em>The rubber band will be stretched 0.02 m.</em>
<em>The work done in stretching is 0.11 J.</em>
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = <em>0.02 m</em>
<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.
potential energy stored = 
==>
= <em>0.11 J</em>
Answer:
ΔT = 
Explanation:
In a simple harmonic motion, specifically in the simple pendulum, the angular velocity
w =
angular velocity and period are related
w = 2π / T
we substitute
2π / T = \sqrt{\frac{g}{L} }
T =
In this exercise indicate that for a long Lo the period is To, then and increase the long
L = L₀ + ΔL
we substitute
T =
T = 
in general the length increments are small ΔL/L «1, let's use a series expansion
we keep the linear term, let's substitute
T =
if we do
T = T₀ + ΔT
T₀ + ΔT =
T₀ + ΔT = T₀ +
ΔT = 