The solution for this problem:
Given:
f1 = 0.89 Hz
f2 = 0.63 Hz
Δm = m2 - m1 = 0.603 kg
The frequency of mass-spring oscillation is:
f = (1/2π)√(k/m)
k = m(2πf)²
Then we know that k is constant for both trials, we have:
k = k
m1(2πf1)² = m2(2πf2)²
m1 = m2(f2/f1)²
m1 = (m1+Δm)(f2/f1)²
m1 = Δm/((f1/f2)²-1)
m 1 = 0.603/
(0.89/0.63)^2 – 1
= 0.609 kg or 0.61kg or 610 g
Answer:
The ball experiences the greater momentum change
Explanation:
The momentum change of each object is given by:
where
m is the mass of the object
v is the final velocity
u is the initial velocity
Both objects have same mass m and same initial velocity u. So we have:
- For the ball, the final velocity is
Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is
- For the clay, the final velocity is
since it sticks to the wall. So, the change in momentum is
So we see that the greater momentum change (in magnitude) is experienced by the ball.
Answer:
Vf = 69.56 cm/s
Explanation:
In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:
s = Vi t + (1/2)at²
where,
s = distance traveled = 160 cm
Vi = Initial Speed = 0 cm/s (since, marble starts from rest)
t = time interval = 4.6 s
a = acceleration = ?
Therefore,
160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²
a = (320 cm)/(4.6 s)²
a = 15.12 cm/s²
Now, we use first equation of motion:
Vf = Vi + at
Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)
<u>Vf = 69.56 cm/s</u>
Hope this helps!!!!!!!!!!!!!
That's called "refraction".
