How many miles would you have to drive for the lower fuel consumption of the second car to compensate for the higher cost of thi
1 answer:
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Answer:
a) a = 6.1 m/s^2
b) a = 0.98m/s^2
Explanation:
Mass of slab = 40kg
Mass of block = 10kg
Coefficient of static friction (Us) = 0.60
Kinetic coefficient (UK) = 0.40
Horizontal force = 100N
The normal reaction from 40kg slab on 10 kg block = 10*9.81
= 98.1N
Static frictional force = Us*R
= 98.1*0.6
= 58.86N
This is less than the force applied
If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force
Kinetic force (Uk*R) = 0.4*98.1
= 39.28N
= 39N
Net force = 100 -39
= 61N
Recall that F = ma
For 10 kg block
a = F/m
a = 61/10
a = 6.1m/s^2
b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4
= 39.24
= 39N
F = ma
a = F/m
For 40kg slab
a = 39/40
a = 0.98m/s^2
Answer:
The burden distance is 7 ft
Solution:
As per the question:
Specific gravity of package emulsion,
Specific gravity of diabase rock,
Diameter of the packaged sticks, d = 3 in
Now,
To calculate the first trail shot burden distance, B:
B = 7 ft