Answer:
distance between the charges is 5.12 × 10⁶ m
Explanation:
charges q₁ = -130.0 C
q₂ = 180 C
force between the charges = 8 N
force between two charge
value of K =8.975 × 10⁹ N.m²/C²
r = 5.12 × 10⁶ m
hence, distance between the charges is 5.12 × 10⁶ m.
Answers:
a)
b)
Explanation:
a) Accoding to the Universal Law of Gravitation we have:
(1)
Where:
is the gravitational force between the eagle and the throng
is the Universal Gravitational constant
is the mass of the eagle
is the mass of the throng
is the distance between the throng and the eagle
(2)
(3) As we can see the gravitational force between the eagle and the throng is quite small.
b) The attraction force between the eagle and Earth is the weight of the eagle, which is given by:
(4)
Where is the acceleration due gravity on Earth
(5)
(6)
Now we can find the ratio between and :
As we can see this ratio is also quite small
The general effort force equation for block and tackle to raise or pull a load can be expressed as
<span><span>S=<span>F/<span>μn</span></span></span>
</span>
where
<span>S
</span> is the effort force,
<span>F
</span> is the load force (often the weight of the object to be moved),
<span>μ
</span> is the mechanical efficiency of the system (1 in an ideal, massless, frictionless system of pulleys), and
<span>n
</span> is the number of ropes between the sets of pulleys.
Answer:
A) I_total = 16 m, B) I_total = 8 m, C) I_total = 8 m, D) I_total = 8 m
Explanation:
The moment of inertia is a scalar quantity, therefore the total moment of inertia
I_total = I₁ + I₂ + I₃ + I₄
the moment of inertia of a point mass with respect to an axis of rotation
I = m r²
Let's apply this to our case
A) Rotation axis at the origin
I₁ = m 0 = 0
for the second masses, we find the distance using the Pythagorean theorem
r =
r = 2 √2
I₂ = m (2 √2) ²
I₂ = 8 m
I₃ = m 2² = 4 m
I₄ = m 2² = 4 m
we substitute
I_total = 0 + 8m + 4m + 4m
I_total = 16 m
B) axis of rotation in the center of the square
let's find the distance to any mass
r =
r = √2
I₁ = m 2
I₂ = m 2
i₃ = m 3
I₄ = m 4
we substitute
I_total = 4 (2m)
I_total = 8 m
C) axis of rotation is the x axis
I₁ = 0
I₂ = m 2² = 4 m
I₃ = m 2² = 4 m
I₄ = 0
I_total = 8 m
D) axis of rotation is the y-axis
I₁ = 0
I₂ = 4m
I₃ = 0
I₄ = 4 m
I_total = 8 m
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