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Jet001 [13]
3 years ago
9

In a collision an object experiences impulses , this impulse can be determined by

Physics
1 answer:
astraxan [27]3 years ago
8 0

In a collision an object experiences impulses, this impulse can be determined by the product of net force exerted upon it in the specific course of time.

<u>Explanation:</u>

An object experiences impulse due to the force exerted upon it in a particular time period.

i.e.

                 \text {Impulse }=F \times \Delta t

Where,

F - Force acted upon the object

\Delta t - time interval for which the force act

According to the definition of Impulse, it is the integral of force (F) that acts upon any object over a time interval ∆t. It produces an equivalent change in the momentum and that too in the same direction as of the applied force (F).

Therefore, in order to find the impulse on an object, we have to find the force acted on it in a specific time interval.

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If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
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3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

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