Answer:
20 m/s
30 m/s
Explanation:
Given:
v₀ = -10 m/s
a = -9.8 m/s²
When t = 1 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (1 s)
v = -19.8 m/s
When t = 2 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (2 s)
v = -29.6 m/s
Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.
Answer:
a) d₁ = 247.8 μm
d₂ = 205.3 μm
b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm
Explanation:
a)
The formula for Michelson Interferometer is derived to be:
d = mλ/2
where,
d = distance moved
m = no. of fringes
λ = wavelength of light
For JAN, we have following data
d = d₁
m = 818
λ = 606 nm = 606 x 10⁻⁹ m
Therefore,
d₁ = (818)(606 x 10⁻⁹ m)/2
<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>
For LINDA, we have following data
d = d₂
m = 818
λ = 502 nm = 502 x 10⁻⁹ m
Therefore,
d₂ = (818)(502 x 10⁻⁹ m)/2
<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>
b)
The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,
Resultant Displacement = Δd = d₁ - d₂
Δd = 247.8 μm - 205.3 μm
<u>Δd = 42.5 μm (in the direction of JAN)</u>
Answer:
order d> a = e> c> b = f
Explanation:
Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form
P₁ = P₂
Using the definition of pressure
F₁ / A₁ = F₂ / A₂
F₂ = A₂ /A₁ F₁
Now we can examine the results
a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2
F₂ = 1.8 / 0.9 4
F₂a = 8 N
b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2
F₂b = 0.45 / 0.9 2
F₂b = 1 N
c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2
F₂c = 3.6 / 1.8 2
F₂c = 4 N
d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2
F₂d = 1.8 / 0.45 4.0
F₂d = 16 m2
e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2
F₂e = 0.9 / 0.45 4
F₂e = 8 N
f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2
F₂f = 0.9 / 1.8 2.0
F₂f = 1 N
Let's classify the structure from highest to lowest
F₂d> F₂a = F₂e> F₂c> F₂b = F₂f
I mean the combinations are
d> a = e> c> b = f
Answer:
velocity = distance / time
Explanation:
Answer:
A. 88,200 J
Explanation:
Given data,
The mass of the crate, m = 200 kg
The height of the shipping container, h = 45 m
The gravitational of a body is possessed by the body due to the virtue of its position.
The formula for gravitational potential energy is,
P.E = mgh joules
Substituting the values
P.E = 200 x 9.8 x 45
= 88,200 J
Hence, the gravitational potential energy of the crate is, P.E = 88,200 J
