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VladimirAG [237]
3 years ago
13

A dragster and driver together have mass 931.9 kg . The dragster, starting from rest, attains a speed of 25.7 m/s in 0.57 s. Fin

d the average acceleration of the dragster during this time interval. Answer in units of m/s 2 .
Physics
1 answer:
Mama L [17]3 years ago
7 0

Answer:

45.09 m/s^2

Explanation:

from the question we were given the following parameters

mass = 931.9 kg

initial velocity (V1) = 0 m/s ( since it started from rest )

final velocity (V2) = 25.7 m/s

time (t) = 0.57 s

The average acceleration is given by:

a = ( V2 - V1 ) / t

therefore

a = (25.7 - 0) / 0.57 = 45.09 m/s^2

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Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

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1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

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∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

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= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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