Answer:
A.) r = 2t
B.) V = 33.5t^3
Explanation:
Given that a spherical balloon is being inflated and the radius of the balloon is increasing at a rate of 2 cm/s
A) Express the radius (r) of the balloon as a function of the time (t).
Since the rate = 2 cm/s that is,
Rate = radius/ time
Therefore,
2 = r/t
Make r the subject of formula
r = 2t
(B) If V is the volume of the balloon as a function of the radius, find V or and interpret it.
Let assume that the balloon is spherical. Volume of a sphere is;
V = 4/3πr^3
Substitute r = 2t into the formula
V = 4/3π(2t)^3
V = 4/3π × 8t^3
V = 32/3 × πt^3
V = 33.5t^3
Answer:
F = 2,894 N
Explanation:
For this exercise let's use Newton's second law
F = m a
The acceleration is centripetal
a = v² / r
Angular and linear variables are related.
v = w r
Let's replace
F = m w² r
The radius r and the length of the rope is related
cos is = r / L
r = L cos tea
Let's replace
F = m w² L cos θ
Let's reduce the magnitudes to the SI system
m = 101.7 g (1 kg / 1000g) = 0.1017 kg
θ = 5 rev (2π rad / rev) = 31,416 rad
w = θ / t
w = 31.416 / 5.1
w = 6.16 rad / s
F = 0.1017 6.16² 0.75 cos θ
F = 2,894 cos θ
The maximum value of F is for θ equal to zero
F = 2,894 N
Answer:
(a) 
(b) 
(c) K.E. = 21.168 J
(d) 
Explanation:
Given:
- mass of a block, M = 3.6 kg
- initial velocity of the block,
- constant downward acceleration,
That a constant upward acceleration of 
is applied in the presence of gravity.
∴
- height through which the block falls, d = 4.2 m
(a)
Force by the cord on the block,
∴Work by the cord on the block,
We take -ve sign because the direction of force and the displacement are opposite to each other.
(b)
Force on the block due to gravity:
∵the gravity is naturally a constant and we cannot change it
∴Work by the gravity on the block,
(c)
Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.
mathematically:
K.E. = 21.168 J
(d)
From the equation of motion:
putting the respective values:
is the speed when the block has fallen 4.2 meters.
C is the correct answer.
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