Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Шада и я тебя понимаю с чего начать с того ни другого человека дал в долг у меня в здоровье
Answer:
decrease the height
Explanation:
height is directly proportional to the g.p.e
The average act on her during the deceleration is 4.47 meters per second.
<u>Explanation</u>:
<u>Given</u>:
youngster mass m = 50.0 kg
She steps off a 1.00 m high platform that is s = 1 meter
She comes to rest in the 10-meter second
<u>To Find</u>:
The average force and momentum
<u>Formulas</u>:
p = m * v
F * Δ t = Δ p
vf^2= vi^2+2as
<u>Solution</u>:
a = 9.8 m/s
vi = 0
vf^2= 0+2(9.8)(1)
vf^2 = 19.6
vf = 4.47 m/s .
Therefore the average force is 4.47 m/s.
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
.............(1)
Now, the second equation of motion is :

Put the value of a in above equation as :




u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.