Answer:
Vx= 11.0865(m/s)
Vy= 6.4008(m/s)
Explanation:
Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:
V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)
The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:
Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)
And for the same theorem the speed on the Y axis will be:
Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)
Answer:
Total mass of combination = 2+3+5 = 10kg.
Acceleration produced = 2m/s^2
hence force =( total mass × acceleration)= (2×10)= 20 N.
Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N
applied force on 2 kg block = 20N
Force between 2 kg and 3 kg block = (20-4) = 16N. ans
Net force on 3 kg block = 3 × 2 =6N.
Applied force on 3 kg block due to 2 kg block = 16N.
hence, force between 3 kg and 5 kg block = (16-6) = 10N .
answers:-
(a) 20 N
(b) 16N
(c) 10 N
Answer:
The maximum amount of work is
Explanation:
From the question we are told that
The temperature of the environment is 
The volume of container A is 
Initially the number of moles is 
The volume of container B is 
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
Now from the Ideal gas law

So substituting for
in the equation above
![W = nRT ln [\frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20nRT%20ln%20%5B%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Where R is the gas constant with a values of 
Substituting values we have that
Answer:
Deltoid Force, 
Additional Information:
Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.
Explanation:
The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.
1. The torque about the point in the shoulder for the deltoid muscle,
2. The torque of the arm,
Assuming the arm is just being stretched and there is no rotation going on,
= 0
= 0
⇒ 

Where,
is radius of the deltoid
is the force of the deltiod
is the angle of the deltiod
is the radius of the arm
is the force of the arm ,
which is the mass of the arm and acceleration due to gravity
is the angle of the arm
The force of the deltoid muscle is,

but
,
∴ 