Given:
u = 7.00m/s, the initial vertical velocity of the package.
Assume that g = 9.8 m/s².
Neglect air resistance.
All quantities are measured as positive upwards.
The height from which the package is dropped is 155 m above ground.
Let t = the time for the package to hit the ground.
Therefore
(7.00 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (- 155 m)
4.9t² - 7t - 155 = 0
or
t² - 1.4286t - 31.6327 0
Solve with the quadratic formula.
t = (1/2)*[1.4286 +/- √(2.0409 + 126.5308)]
= 6.384 s or - 4.955 s
Reject negative time.
Answer: 6.384 s
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be .
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ =
= 1.63 m/s²
From the question, T=2π√L/g
We can write that,
.......... (1)
Where is the period of the pendulum on Earth and is the measure of the strength of Earth's gravity
and
.......... (2)
Where is the period of the pendulum on Moon and is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, and are constants.
Dividing equation (1) by (2), we get
From the question,
= 9.8 m/s²
= 1.63 m/s²
= ??
From,
∴
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Because the gravity pulls use towards the ground and our bones can’t take the amount of weight hitting the surface
Answer:
(a) W= 44N
(b)W= 31.65 N
Explanation:
Data
T=44 N : Maximum force that the rope can withstand without breaking
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) We apply the formula (1) at constant speed , then, a=0
W: heaviest fish that can be pulled up vertically
∑F = 0
T-W =0
W = T
W= 44N
(b) We apply the formula (1) , a= 1.26 m/s²
W: heaviest fish that can be pulled up vertically
W= m*g
m= W/g
g= 9.8 m/s² : acceleration due to gravity
∑F = 0
T-W = m*a
T= W+(W/g)*a
44=W*(1+1/9.8)* (1.26 )
44= W* 1.39
W= 44/1.39
W= 31.65 N