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Nutka1998 [239]
3 years ago
12

A train is moving with a velocity of 20 m/s. when the brakes are applied the acceleration is reduced to -0.6 m/s².calculate the

distance covered by the train before coming to rest​
Physics
1 answer:
Serhud [2]3 years ago
3 0

Answer:

x = 333.33 [m]

Explanation:

To solve this problem we must use the following kinematics equation.

v_{f}^{2} = v_{i}^{2}-(2*a*x)

where:

Vf = final velocity = 0

Vi = initial velocity = 20 [m/s]

a = desacceleration = 0.6 [m/s^2]

x = distance [m]

Note: the final speed is zero as the body finishes its movement.

Now replacing:

0 = (20)^2 - (2*0.6*x)

1.2*x = 400

x = 333.33 [m]

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1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche
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Answer:

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(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2

A^2=\dfrac{mv^2}{k}

Put the value into the relation

A^2=\dfrac{1.60\times13.0^2}{315}

A=\sqrt{0.858}

A=0.926\ m

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

a=A\omega^2

a=A\times\dfrac{k}{m}

Put the value into the formula

a=0.926\times\dfrac{315}{1.60}

a=182.31\ m/s^2

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

F=ma

Put the value into the formula

F= 1.60\times182.31

F=291.69\ N

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat
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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

                               j_{H_2O}=\frac{D}{l} \bigtriangleup c

where

  • j_{H_2O} is the mass flux of the water which is to be calculated.
  • D is diffusion coefficient which is given as 0.25 cm^2/s
  • l is the thickness of the film which is 0.15 cm thick.
  • \bigtriangleup c is given as

                                \bigtriangleup c= \frac{P_{sat}-P_a}{RT}

In this

  • P_{sat} is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa
  • P_a is the air pressure which is given as 0.5 times of P_{sat}
  • R is the universal gas constant as 8.314 kJ/kmol-K
  • T is the temperature in Kelvin scale which is 20+273= 293K

By substituting values in the equation

                                    \bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3

Converting \bigtriangleup c into cm^3/cm^3

As 1 mole of water 18 cm^3 so

                               \bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6}  cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6}  cm^3/cm^3

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                            j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6}  cm/s

For calculation of water level drop in a day, converting mass flux as

                     j_{H_2O}=14.4 \times 10^{-6}  \times 24 \times 3600  cm/day\\ j_{H_2O}=1.24  cm/day

So the water level will drop by about 1.24 cm in 1 day.

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3 years ago
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