Okay. There is a simple formula to go with this where:
p = mv
P: Momentum.
M: Mass.
V: Velocity
Sub the numbers in and solve for M.
10.0 = m(1.5)
10.0/1.5 = m
6.67 kg = m
Therefore the mass of the ball is 6.67kg.
Every cell has chromosomes in its nucleus. In a human cell,
there are 26 of them. AND ... lined up on every chromosome
are thousands and thousands of GENES. Those carry the
instructions for what kind of organ or bone or tissue or hair
this cell will be used to build, how the structure will work, how
tall you will be, what color your eyes will be, how deep your
voice will be, and what illnesses you will have.
Every cell has the complete set of instructions in it. In the past
several years, we have just started to be able to read them, and
extraordinary progress has been made. There's an awful long way
still left to go.
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U =
................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 = 
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U = 
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J
Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:

Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!