Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
Answer:
B. 
Explanation:
Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:
(1)
(2)
Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because
,
and R is an universal constant:
, solving for 


The spring constant is 181.0 N/m
Explanation:
We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where
k is the spring constant of the spring
x = 2.08 cm = 0.0208 m is the compression of the spring
m = 12.3 g = 0.00123 kg is the mass of the dart
is the acceleration due to gravity
h = 3.25 m is the maximum height of the dart
Solving for k, we find:

Learn more about potential energy:
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To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos <span>θ
csc </span>θ = 1 / sin θ<span>
Using these identities:
</span>cot θ ∙ sec θ = (cos θ / sin θ ) (<span> 1 / cos </span><span>θ)
</span>
We can cancel out cos <span>θ, leaving us with
</span>cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc <span>θ</span>
Answer:

Explanation:
As we know that Far sighted person has near point shifted to 80 cm distance
so he is able to see the object 80 cm
now the distance of lens from eye is 2 cm
and the person want to see the objects at distance 10 cm
so here the image distance from lens is 80 cm and the object distance from lens is 8 cm
now from lens formula we have


