Answer:
Tension in the string will increase
Explanation:
As we know that tension in the string at any angle with the vertical is given as

now we have

also we know that
angular speed of the stone is directly depending on the time period of the motion
so it is given as

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s
so the angular speed would be doubled
So here we can say that
tension in the string will increase when we will increase the frequency of revolution.
Answer:
457.81 Hz
Explanation:
From the question, it is stated that it is a question under Doppler effect.
As a result, we use this form
fo = (c + vo) / (c - vs) × fs
fo = observed frequency by observer =?
c = speed of sound = 332 m/s
vo = velocity of observer relative to source = 45 m/s
vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).
fs = frequency of sound wave by source = 459 Hz
By substituting the the values to the equation, we have
fo = (332 + 45) / (332 - (-46)) × 459
fo = (377/ 332 + 46) × 459
fo = (377/ 378) × 459
fo = 0.9974 × 459
fo = 457.81 Hz
Ignoring fluid resistance, football will <span>maintain a constant speed until other forces accelerate the football.</span>
Answer:
232.641374 mph
Explanation:
A race car has a maximum speed of 0.104km/s
Let X represent the speed in miles per hour
Therefore the speed in miles per hour can be calculated as follows
1 km/s = 2,236.936292 mph
0.104km/s = X
X = 0.104 × 2,236.936292
X = 232.641374
Hence the speed in miles per hour is 232.641374 mph
Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:

Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for ![\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}](https://tex.z-dn.net/?f=%5Comega%5B%5Ctex%5D%20and%200.8%20m%20for%20l%20%5Btex%5D%5Cbegin%7Barray%7D%7Bc%7D%5C%5CL%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7Dm%7Bl%5E2%7D%5Comega%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7D%5Cleft%28%20%7B0.120%7B%5Crm%7B%20kg%7D%7D%7D%20%5Cright%29%7B%5Cleft%28%20%7B%7B%5Crm%7B80%7D%7D%7B%5Crm%7B.0%20cm%7D%7D%7D%20%5Cright%29%5E2%7D%7B%5Cleft%28%20%7B%5Cfrac%7B%7B1%20%5Ctimes%20%7B%7B10%7D%5E%7B%20-%202%7D%7D%7B%5Crm%7Bm%7D%7D%7D%7D%7B%7B1%7B%5Crm%7B%20cm%7D%7D%7D%7D%7D%20%5Cright%29%5E2%7D%5Cleft%28%20%7B%7B%5Crm%7B3%7D%7D%7B%5Crm%7B.00%20rad%2Fs%7D%7D%7D%20%5Cright%29%5C%5C%5C%5C%20%3D%200.0192%7B%5Crm%7B%20kg%7D%7D%20%5Ccdot%20%7B%7B%5Crm%7Bm%7D%7D%5E%7B%5Crm%7B2%7D%7D%7D%7B%5Crm%7B%2Fs%7D%7D%5C%5C%5Cend%7Barray%7D)