Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
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: 0.392 m/s
Explanation:
The Doppler shift equation is:
Where:
is the actual frequency of the sound wave
is the "observed" frequency
is the speed of sound
is the velocity of the observer, which is stationary
is the velocity of the source, which are the red blood cells
Isolating :
Finally:
Here you go it was too long to type
