True or false as heat and pressure decreases one type of metamorphic rock will transform into another type
1 answer:
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Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Answer:
0.000001 kg
Explanation:
because 1 kg equal 1,000,000 milligrams
we take which equals 0.000001 kg
Answer:
the work done to lift the bucket = 3491 Joules
Explanation:
Given:
Mass of bucket = 10kg
distance the bucket is lifted = height = 11m
Weight of rope= 0.9kg/m
g= 9.8m/s²
initial mass of water = 33kg
x = height in meters above the ground
Let W = work
Using riemann sum:
the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)
=
Work done in lifting the bucket (W) = force × distance
Force (F) = mass × acceleration due to gravity
Force = 9.8 * 10 = 98N
W done by bucket = 98×11 = 1078 Joules
Work done to lift the rope:
At Height of x meters (0≤x≤11)
Mass of rope = weight of rope × change in distance
= 0.8kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0
W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]
=7.84(60.5 -0) = 474.32 Joules
Work done in lifting the water
At Height of x meters (0≤x≤11)
Rate of water leakage = 36kg ÷ 11m = kg/m
Mass of rope = weight of rope × change in distance
= kg/m × (11-x)m = 3.27kg/m × (11-x)m
W done = integral of (mass×g ×distance) with upper 11 and lower limit 0
W done =
Note : upper limit is 11 not 1, problem with math editor
W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0
W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]
= 32.046(60.5 -0) = 1938.783 Joules
the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)
the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103
the work done to lift the bucket = 3491 Joules