What is 5C in physics
1 answer:
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Answer:
a)
b)
c)
d)
e) &
f)
Explanation:
From the question we are told that:
Stretch Length
Mass
Total stretch length
a)
Generally the equation for Force F on the spring is mathematically given by
b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by
Where
A=Amplitude
And
Therefore
c)
Generally the equation for Max Acceleration of Mass on the spring is mathematically given by
d)
Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by
e)
Generally the equation for the period T is mathematically given by
Generally the equation for the Frequency is mathematically given by
f)
Generally the Equation of time-dependent vertical position of the mass is mathematically given by
Where
'= signify order of differentiation
Answer:
a) perfectly inelastic, b) collision is inelastic, c) elastic
Explanation:
In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved
Initial instant. Before the skater touches the frisbee
p₀ = M v₁ + m v₂
where M and m are the masses of the skater and frisbee, respectively
for the final moment they give us several possibilities, in all case the moment is conserved
p₀ =
case a)
Final instant. grabs the frisbee and holds it
p_{f} = (M + m) v '
p₀ = p_{f}
We can see that this shock is perfectly inelastic, it holds the fressbee
case b)
final instant.
This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.
case c)
final instant. Grab the fressbee and resend it
this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut + a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s = a
= x 14 x
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.