Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
The magnitude of the torque on the loop due to the magnetic field is 
.
Explanation:
Given that,
Diameter = 10 cm
Current = 0.20 A
Magnetic field = 0.30 T
Unit vector
We need to calculate the torque on the loop
Using formula of torque
Where, N = number of turns
A = area
I = current
B = magnetic field
Put the value into the formula
Hence, The magnitude of the torque on the loop due to the magnetic field is .
To
solve this problem, we assume that the wavelength of the light in air is 500
nanometers.
For this case we
only need the refractive index of the polystyrene. For an antireflective
coating, we need a quarter of wave thickness at the wavelength in the air. Which
means that the antireflective coating needs to be as thick as 1/4 of the
wavelength, divided by the coating’s refractive index. This is expressed
mathematically in the form:
x = λ / (4 * n)
where,
x = thickness
λ = wavelength
of light
n = index of
refraction of polystyrene
Substituting:
x = 500 nm / (4
* 1.49)
x = 500 nm / 5.96
x = 83.90 nm
