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leonid [27]
2 years ago
10

Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the

surroundings, calculate the final temperature, T f , of the water after all the ice melts.
Physics
1 answer:
Lelechka [254]2 years ago
7 0

Answer:

Ft = 17.48°C

Explanation:

Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

Therefore,

Heat absorbed by water of 20g = heat rejected by water of 265g.

So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)

So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)

To get;

7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

So factorizing, we get;

83.74 Ft + 1193.3 Ft = 29832.4 - 7513

So; 1277.04 Ft = 22319.4

So; Ft = 22319.4/1277.04 = 17.48°C

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A steel cable has a cross-sectional area 2.54 10-3 m2 and is kept under a tension of 1.01 104 N. The density of steel is 7860 kg
ElenaW [278]

Answer:

The speed is equals to 22.49 m/s

Explanation:

Given Data:

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Solution:

As we know that

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Now the equation for speed of traverse wave is calculated through:

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A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260
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Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

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Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

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What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

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Where, I = moment of inertia

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Here, F = tension

The force is

F=m(g-a)...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

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a=\dfrac{g}{1+\dfrac{M}{2m}}

Put the value into the formula

a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}

a=6.84\ m/s^2

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Using equation (I)

F=\dfrac{M}{2}a

Put the value into the formula

F=\dfrac{12.1}{2}\times6.84

F=41.38\ N

Hence, The tension in the rope is 41.38 N.

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