Answer:
-0.0047 rad/s²
335.103 seconds
99.18 seconds
Explanation:
= Final angular velocity
= Initial angular velocity = 1.5 ra/s
= Angular acceleration
= Angle of rotation = 40 rev
t = Time taken
Equation of rotational motion

Acceleration while slowing down is -0.0047 rad/s²

Time taken to slow down is 335.103 seconds

Solving the equation

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103
Horizontal and Vertical are the two motions that combine to produce projectile motion.
Answer:
From the question we are told that
The length of the rod is 
The speed is v
The angle made by the rod is 
Generally the x-component of the rod's length is

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

=> ![L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }](https://tex.z-dn.net/?f=L_xo%20%20%3D%20%20%5BL_o%20cos%20%28%5Ctheta%20%29%5D%20%20%5Csqrt%7B1%20%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D)
Generally the y-component of the rods length is mathematically represented as

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e
Generally the resultant length of the rod as seen by the observer is mathematically represented as

=> ![L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}](https://tex.z-dn.net/?f=L_r%20%20%3D%20%5Csqrt%7B%5B%20%28L_o%20cos%28%5Ctheta%29%20%5B%5Csqrt%7B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%5C%20%5C%20%5D%5E2%2B%20L_o%20sin%28%5Ctheta%20%29%5E2%29%7D)
=> ![L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}](https://tex.z-dn.net/?f=L_r%3D%20%5Csqrt%7B%20%28L_o%20cos%28%5Ctheta%29%5E2%20%2A%20%5B%20%5Csqrt%7B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%20%5D%5E2%20%2B%20%28L_o%20sin%28%5Ctheta%29%29%5E2%7D)
=> ![L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}](https://tex.z-dn.net/?f=L_r%20%20%3D%20%5Csqrt%7B%28L_o%20cos%28%5Ctheta%29%20%5E2%20%5B1%20-%20%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%5D%20%2B%28L_o%20sin%28%5Ctheta%29%29%5E2%7D)
=> ![L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}](https://tex.z-dn.net/?f=L_r%20%3D%20%20%5Csqrt%7BL_o%5E2%20%2A%20cos%5E2%28%5Ctheta%29%20%20%5B1%20-%20%5Cfrac%7Bv%5E2%20%7D%7Bc%5E2%7D%20%5D%2B%20L_o%5E2%20%2A%20sin%28%5Ctheta%29%5E2%7D)
=> ![L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }](https://tex.z-dn.net/?f=L_r%20%20%3D%20%20%5Csqrt%7B%20%5Bcos%5E2%5Ctheta%20%2Bsin%5E2%5Ctheta%20%5D-%20%5Cfrac%7Bv%5E2%20%7D%7Bc%5E2%7Dcos%5E2%20%5Ctheta%20%7D)
=> 
Hence the length of the rod as measured by a stationary observer is

Generally the angle made is mathematically represented

=> 
=>
Explanation:
Answer:
V initial = 29.4 m.s²
Explanation:
( Using the laws of motion)
V final = V initial + Acceleration × time
0 = V initial + ( -9.8)(3)
29.4 = V initial
* I took upward as positive that's why I substituted -9.8 *
* for V final we know that at maximum height the ball is not moving thats why is = 0 *
Answer:
T_final = 279.4 [°C]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.

where:
Q = heat = 9457 [cal]
m = mass = 79 [g] = 0.079 [kg]
Cp = specific heat = 0.5 [cal/g*°C]
T_initial = initial temperature = 40 [°C]
T_final = final temperature [°C]
![9457 = 79*0.5*(T_{f}-40)\\239.41=T_{f}-40\\\\T_{f}=279.4[C]](https://tex.z-dn.net/?f=9457%20%3D%2079%2A0.5%2A%28T_%7Bf%7D-40%29%5C%5C239.41%3DT_%7Bf%7D-40%5C%5C%5C%5CT_%7Bf%7D%3D279.4%5BC%5D)