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snow_tiger [21]
3 years ago
8

A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a

rough patch in the ice which slows it down to a speed of 2.35 m/s. How much energy is dissipated as the puck slides over the rough patch
Physics
1 answer:
Anton [14]3 years ago
5 0

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

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To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

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E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}

where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

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