Arrow at the left side pointing towards right side represents the frictional force as it always acts opposite to motion
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sum of Mc = 0 = -Ay(4.2 + 3cos(59)) + (275)(2.1 + 3cos(59)) + M
<span>- Ay = (M + (275*(2.1 + 3cos(59)))/(4.2 + 3cos(59)) </span>
<span>sum of Ma = 0 = (-275)(2.1) - Cy(4.2 + 3cos(59)) + M </span>
<span>- Cy = (M - (275*2.1))/(4.2 + 3cos(59)) </span>
<span>Ay + Cy = 275 = ((M+1002.41)+(M-577.5))/(5.745) </span>
<span>= (2M + 424.91)/(5.745) </span>
<span>M = ((275*5.745) - 424.91)/2 </span>
<span>= 577.483 which rounds off to 577 </span>
<span>Is it maybe supposed to be Ay - Cy = 275</span>
Answer:
The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

Explanation:
From the question we are told that
The mass of the rocket is m = 6300 kg
The velocity at gas is being ejected is u = 2000 m/s
The initial acceleration desired is 
The time taken for the gas to be ejected is t = 1 s
Generally this desired acceleration is mathematically represented as

Here
is the rate at which gas is being ejected with respect to time
Substituting values

=> 
=> 
=> 
=> 
Answer:
t = 25 seconds
Explanation:
Given that,
Distance, d = 115 m
Initial speed, u = 4.2 m/s
Final speed, v = 5 m/s
We need to find the time taken in increasing the speed.
We know that,
Acceleration,
....(1)
The third equation of kinematics is as follows :

Hence, it will take 25 seconds to increase the speed.
Explanation:
(b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.