Answer:
a) 37171
b) 
sec
Explanation:
m = mass of the weight being lifted = 60.0 kg
d = distance by which the weight is lifted = 0.670 m
E = Energy available to burn = 1 lb = 3500 kcal = 3500 x 4184 J
g = acceleration due to gravity = 9.8 m/s²
n = number of times the weight is lifted
Energy available to burn is given as
E = n m g d
3500 x 4184 = n (60) (9.8) (0.670)
n = 37171
b)
T = time period for each lift up = 1.80 s
t = total time taken
Total time taken is given as
sec
Chemical Reaction between metal oxide and water solution
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
<h3>Force exerted outside the wheel</h3>
The force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
- Let the distance traveled by the load = 1.5 m
- Let the radius of the wheel or position of the force = 0.45 m
∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Learn more about angular momentum here: brainly.com/question/7538238
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