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cupoosta [38]
3 years ago
8

4.) What are examples of physical and chemical properties?

Physics
1 answer:
lubasha [3.4K]3 years ago
7 0

Answer:

Some physical properties are odor (Sewage smell), and melting and boiling point (boiling water melting ice).

Some chemical properties are  flammability (burning wood), and oxidation (rotting apple).

Explanation:

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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial
nalin [4]

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)

= \int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx

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 = \int\limits^1 _0 {9.8*0.8(11-x)} \, dx

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 = \frac{36}{11}kg/m

Mass of rope = weight of rope × change in distance

= \frac{36}{11}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 = \int\limits^1 _0 {9.8*3.27(11-x)} \, dx

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

8 0
3 years ago
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