A capacitor with air between its plates ischarged to 60 V and then disconnected fromthe battery. When a piece of glass is placed
1 answer:
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Answer:
The period of motion of new mass T = 0.637 sec
Explanation:
Given data
Mass of object (m) = 9 gm = 0.009 kg
Δx = 3.5 cm = 0.035 m
We know that spring force is given by
F = m g
F = 0.009 × 9.81 = 0.08829 N
Spring constant
k = 2.522
New mass = 26 gm = 0.026 kg
Now the period of motion is given by
T = 0.637 sec
This is the period of motion of new mass.
Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant;
Age of the universe;
We know that, the reciprocal of the Hubble's constant ( ) gives an estimate of the age of the universe (
). It is expressed as:
Now,
Hubble's constant;
We know that;
so
Therefore;
Now, we input this Hubble's constant value into our equation;
Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
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Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d= * tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d= * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip