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svetoff [14.1K]
3 years ago
15

4. In the Exploration, the sled does not move at all when two equal forces are applied

Physics
1 answer:
Assoli18 [71]3 years ago
7 0

Answer:

The net force is zero, so the acceleration is zero

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force applied to it, according to the equation:

\sum F = ma (1)

where

\sum F is the net force on the object

m is the mass of the object

a is its acceleration

In this problem, we have a sled acted upon two forces, F_1, F_2. So the net force on the sled is

\sum F = F_1 + F_2 (2)

however, we are told that the two forces are equal in magnitude but in opposite directions, so

F_1 = F\\F_2 = -F

So, eq.(2) becomes

\sum F = F+(-F) = 0

and so eq.(1) becomes

\sum F = ma = 0

which means

a=0

so the acceleration of the sled is zero, and if the sled was at rest, it will not move.

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5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.​
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Answer: 0.8 g/cm

Explanation:

p= m/V

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write down the total momentum for two marbles of mass,m, both moving at velocity, v. What is the kinetic energy of the system.
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An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

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>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

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

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

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

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 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

8 0
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A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t
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Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

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now,

if the efficiency is equal to 42$ = 0.42

 T_C= (1 - 0.42) \times 464

 T_C= 0.58 \times 464

 T_C= 269.12\ K

8 0
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