how fast is a ball going when it hits the ground after being dropped from a height of 16m the acceleration of gravity is 9.8
1 answer:
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Answer:
Therefore, the force supported by the hydraulic cylinder is = -70.01 lb
Thus, the force supported by the bar = -233.84 lb
The reaction force supported by the reaction of the roller = 341.31 lb
Explanation:
The attached diagrams is shown in the file below:
From there; taking moments about point A
- 40 (180) - (80) Cos 37° + 55
-7200 + (-80 Cos 37° + 55 sin 37° ) = 0
= - 7200 - 30.79
- 30.79 = 7200
=
= -233.84 lb
Thus, the force supported by the bar = -233.84 lb
Taking the equilibrium of forces in the vertical direction
sin 37 +
cos 20 -
= 0
- 233.84 sin 37 + cos 20 - 180 = 0
cos 20 = 320.73
=
= 341.31 lb
The reaction force supported by the reaction of the roller = 341.31 lb
Taking the equilibrium of forces on the horizontal direction.
cos 37 -
sin 20° = 0
-233.84 cos 37 - + 341.31 sin 20° = 0
-186.75 - + 116.74 = 0
- -70.01 = 0
- = 70.01
= - 70.01 lb
Therefore, the force supported by the hydraulic cylinder is = -70.01 lb
119.7kPa
Explanation:
Given parameters:
Pressure of gas in balloon = 111kPa
Temperature of gas = 22°C
Final temperature = 45°C
Unknown:
Final pressure = ?
Solution:
Since the gases in the balloon have the same number of moles. We can apply a derivative of the combined gas law to solve this problem.
At constant volume the pressure of a given mass of gas varies directly with the absolute temperature.
P1 is the initial pressure
P2 is the final pressure
T1 is the initial temperature
T2 is the final temperature
convert from celcius to kelvin:
tK = 273 + tC
T1 = 273 + 22 = 295K
T2 = 273 + 45 = 318K
P2 = 119.7kPa
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