Explanation:
The ball is in free fall (gravity is the only force acting on the ball), so its acceleration is 9.8 m/s² down during the entire path.
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
K.E = 1/2 mv²
800 = 1/2 ×12 ×v²
800 = 6 v²
800 / 6 = v²
= 133.4 =v²
√133.4 = √v²
11.5 = v²
I hope this answer is correct.
Answer:
a) fr = 224.3 N
, b) fr = 224.3 N
, c) v = 198.0 m/s
Explanation:
a) For this exercise let's start by calculating the acceleration in the fall
v² = v₀² - 2 a (y-y₀)
When it jumps the initial vertical speed is zero
a = -v² / 2 (y-y₀)
a = -68 2/2 (1000-2000)
a = 2,312 m / s²
Let's use the second net law to enter the average friction force
fr = m a
fr = 97 2,312
fr = 224.3 N
b) let's look for acceleration
v² = v₀² - 2 a y
a = (v² –v₀²) / 2 (y-y₀)
a = (4² - 68²) / 2 (0-1000)
a = 2,304 m / s²
fr = m a
fr = 97 2,304
fr = 223.5 N
c) the speed of the wallet is searched with kinematics
v² = v₀² - 2 g (y-y₀)
v = √ (0-2 9.8 (0-2000))
v = 198.0 m/s
Downward movement under the force of gravity only.
