Answer:
a. 12.12°
b. 412.04 N
Explanation:
Along vertical axis, the equation can be written as
T_1 sin14 + T_2sinA = mg
T_2sinA = mg - T_1sin12.5 ....................... (a)
Along horizontal axis, the equation can be written as
T_2×cosA = T_1×cos12.5 ......................... (b)
(a)/(b) given us
Tan A = (mg - T_1sin12.5) / T_1 cos12.5
= (176 - 413sin12.5) / 413×cos12.5
A = 12.12 °
(b) T2 cosA = T1 cos12.5
T2 = 413cos12.5/cos12.12
= 412.04 N
Answer:
Average acceleration is 
Explanation:
It is given that,
Initial velocity, u = 0
Final velocity, v = 6.5 km/s = 6500 m/s
Time taken, t = 60 s
Acceleration, 

Since, 
So, 
So, the angular acceleration of the missile is
. Hence, this is the required solution.
Answer:
D
Explanation:
The decrease in potential energy is equal to the increase in kinetic energy.
mgh
250 x 9.8 x 30
=73, 500
Complete Question
A football coach walks 18 meters westward, then 12 meters
eastward, then 28 meters westward, and finally 14 meters
eastward.
a
From this motion what is the distance covered
b
What is the magnitude and direction of the displacement
Answer:
a

b
Magnitude
Direction
West
Explanation:
From the question we are told that
The first distance covered westward is ![d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2](https://tex.z-dn.net/?f=d_w_1%20%20%3D%20%2018%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20The%20%20first%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e1%20%3D%20%2012%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20second%20distance%20covered%20westward%20is%20%5Btex%5Dd_w_2%20%20%3D%20%2028%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20%20second%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e2%20%3D%20%2014%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%3C%2Fp%3E%3Cp%3EGenerally%20the%20distance%20covered%20is%20mathematically%20represented%20as%20%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%5Btex%5DD%20%3D%20%20d_w1%20%2B%20d_w2%20%2B%20d_e1%20%2B%20d_e2)
=> 
=> 
For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative
The magnitude of the displacement is

=>
=>
The direction is west
Answer:
KE = KE (incidental) - KE of emitted photons
or KE = h * f - Wf
So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy
If you double the frequency then h * f = 6.16
KE = 6.16 - 1.2 = 4.96 eV