The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
Answer:
I(x) = 1444×k ×
I(y) = 1444×k ×
I(o) = 3888×k ×
Explanation:
Given data
function = x^2 + y^2 ≤ 36
function = x^2 + y^2 ≤ 6^2
to find out
the moments of inertia Ix, Iy, Io
solution
first we consider the polar coordinate (a,θ)
and polar is directly proportional to a²
so p = k × a²
so that
x = a cosθ
y = a sinθ
dA = adθda
so
I(x) = ∫y²pdA
take limit 0 to 6 for a and o to 
for θ
I(x) = 
y²p dA
I(x) = (a sinθ)²(k × a²) adθda
I(x) = k 
da × 
(sin²θ)dθ
I(x) = k da × (1-cos2θ)/2 dθ
I(x) = k 
× 
I(x) = k × 
× ( 
I(x) = k × 
× 
I(x) = 1444×k × .....................1
and we can say I(x) = I(y) by the symmetry rule
and here I(o) will be I(x) + I(y) i.e
I(o) = 2 × 1444×k ×
I(o) = 3888×k × ......................2
Answer:
No
Explanation:
She will not be able to measure the length of her window accurately due to instrumental error from her choice of instrument. The elastic nature of her tape would alter the measurement because it will stretch as she is taking her readings, thus reducing the true measurement of the length of her window.
To measure the length of her window, she could use an inelastic tape rule or a metre rule. These instruments would eliminate instrumental error.
Answer:
1.170*10^-3 m
3.23*10^-32 m
Explanation:
To solve this, we apply Heisenberg's uncertainty principle.
the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
If we make Δx the subject of formula, by rearranging, we have
Δx = h / 4π * m(e).Δv
on substituting the values, we have
for the electron
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 5.67*10^-31
Δx = 1.170*10^-3 m
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2
Δx = 6.63*10^-34 / 0.021
Δx = 3.23*10^-32 m
therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet
c. energy and object has.....
