Answer:
59cm
Explanation:
angular velocity = 0.8 rad/s
linear velocity = angular velocity * radius
=0.8rad/s * 5m
= 4 m/s
wavelength = (V + U)/F
where,
V is the velocity of the wave
U is the velocity of the source
F is the frequency of the source.
wavelength = (350 m/s + 4 m/s ) / 600 Hz
Wavelength = 0.59m or 59 cm
The labeled points which is Letter B in the given Image is the point that the axis of rotation passes through. This problem is an example of rotational dynamics, formerly an object moves in a straight line then the motion is translational but when an object at inactivity lean towards to continue at inactivity and an object in rotation be possible to continue rotating with continuous angular velocity unless bound by a net external torque to act then is rotational. In a rotational motion, the entity is not treated as a constituent part but is treated in translational motion. It points out with the study of torque that outcomes angular accelerations of the object.
Answer:
Exploratory
Explanation:
<u>Focus groups</u>
It is a small group of 8-12 respondents guided by a moderator through a thorough debate on a specific subject or idea.It's great for generation of ideas, brainstorming, insight into motives, attitudes, and perceptions. it can show likes, dislikes,emotional requirements and prejudices.
Exploratory methods are used to gain initial insights that could pave the way for further investigation.
Some of the exploratory methods are focus groups, Key informant,case studies,secondary data and observational data.
The equation to be used is written as:
ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol
r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³
ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³
Complete Question
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?
Answer:
The kinetic energy is 
Explanation:
From the question we are told that
The mass of the block is 
The spring constant is 
The length of first displacement is 
The length of first displacement is 
At the
the kinetic energy is mathematically evaluated as

Where
is the change in energy stored on the spring which is mathematically represented as

=> 
Substituting value

