<h3><u>Answer;</u></h3>
Radius = 0.0818 m
Angular velocity = 2.775 × 10^7 rad/sec
<h3><u>Explanation;</u></h3>
The mass of proton m=1.6748 × 10^-27 kg;
Charge of electron e= 1.602 × 10^-19 C;
kinetic energy E= 2.7 MeV
= 2.7 × 10^6 × 1.602 × 10^-19 J;
= 4.32 × 10^-13 Joules
But; K.E =0.5m*v^2,
Hence v=√(2K.E/m)
Velocity = 2.27 × 10^7 m/s
Angular velocity, ω = v/r
Therefore; V = ωr
Hence; V = √(2K.E/m) = ωr
r= √(2E/m)/w = √E*√(2*m)/(eB)
= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)
but E = 4.32 × 10^-13 Joules
r = 0.0818 m
Angular speed
Angular velocity, ω = v/r , where r is the radius and v is the velocity
Therefore;
Angular velocity = 2.27 × 10^7 / 0.0818 m
= 2.775 × 10^7 rad /sec
The answer should be d because they are constantly rotating
Answer:30 m
Explanation:
Given
Maximum Horizontal distance is 5 m on earth
launching angle
Acceleration due to gravity on earth is 
Acceleration due to gravity on moon is 
Range of projectile is given by

----1
-----2
Divide 1 & 2


The mass of the aeroplane is 300,000 kg.
<h3>What is Newton's second law of motion?</h3>
It states that the force F is directly proportional to the acceleration a of the body and its mass.
The law is represented as
F =ma
where acceleration a = velocity change v / time interval t
Given is the aeroplane lands at a speed of 80 m/s. After landing, the aeroplane takes 28 s to decelerate to a speed of 10 m/s. The mean resultant force on the aeroplane as it decelerates is 750 000 N.
The force expression will be
F = mv/t
Substitute the values and we have
750000 = m x (80 -10)/ 28
750,000 = m x 2.5
m = 300,000 kg
Thus, the mass of the aeroplane is 300,000 kg.
Learn more about Newton's second law of motion.
brainly.com/question/13447525
#SPJ1
To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.
The linear mass density is given as,



The expression for the wavelength of the standing wave for the second overtone is

Replacing we have


The frequency of the sound wave is



Now the velocity of the wave would be



The expression that relates the velocity of the wave, tension on the string and linear mass density is





The tension in the string is 547N
PART B) The relation between the fundamental frequency and the
harmonic frequency is

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

Then,

Rearranging to find the fundamental frequency


