Answer:
f = 878,080 N
Explanation:
mass of pile driver (m) = 2100 kg
distance of pile driver to steel beam (s) = 5 m
depth of steel driven (d) = 12 cm = 0.12 m
acceleration due to gravity (g0 = 9.8 m/s^{2}
calculate the average force exerted on the pile driver by the beam.
- from work done = force x distance
- work done = change in potential energy of the pile driver
- equating the two equations above we have
force x distance = m x g x (s - d)
f x 0.12 = 2100 x 9.8 x (5- (-0.12))
d = - 0.12 because the steel beam went down at we are taking its
initial position to be an origin point which is 0
f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12
f = 878,080 N
F-free = m*g - F_air = m*a
F_air = 1.2 * m
a= (105 kg * 9.8 m.s^2 - 5*105) / 105 kg
a = 9.3 m/s
Hope this helps
Answer:
c. about 1/10 as great.
Explanation:
While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.
This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.
Mathematically:


since mass is constant

when 
then,


the body will experience the tenth part of the maximum force.
where:
represents the rate of change in dependent quantity with respect to time
momentum
mass of the person jumping
velocity of the body while hitting the ground.
<h2>
Answer: 12 s</h2>
Explanation:
The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.
In this sense, the main movement equation in the Y axis is:
(1)
Where:
is the instrument's final position
is the instrument's initial position
is the instrument's initial velocity
is the time the parabolic movement lasts
is the acceleration due to gravity at the surface of planet X.
As we know
and
when the object hits the ground, equation (1) is rewritten as:
(2)
Finding
:
(3)
(4)
(5)
Finally:

Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
<h3>At the bottom condition</h3>
T - m g = m a
T = m ω² r + m g
<h3>At the top condition</h3>
T + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.