Answer:
|v| = 8.7 cm/s
Explanation:
given:
mass m = 4 kg
spring constant k = 1 N/cm = 100 N/m
at time t = 0:
amplitude A = 0.02m
unknown: velocity v at position y = 0.01 m
1. Finding Ф from the initial conditions:
2. Finding time t at position y = 1 cm:
3. Find velocity v at time t from equation 2:
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:
Therefore the horizontal range = 294897.96 m.
Explanation:
Range of a projectile: The range is defined as the horizontal distance from the point of projection to the point where the projectile hit the projection plane again. The S.I unit of range is Meter (m).
It can be expressed mathematically as
R = u²sin2∅/g............................. Equation 1
Where R = Horizontal range, ∅ = angle of projection, u = initial velocity, g = acceleration due to gravity.
<em>Given: u = 1700 m/s, </em>∅ = 55°,
Constant: g = 9.8 m/s²
Substituting these values into equation 1
R = (1700²sin55)/9.8
R = 2890000/9.8
R = 294897.96 m.
Therefore the horizontal range = 294897.96 m.
