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3 years ago

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A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

ar.? b) the distance covered by the car

1 answer:

stealth61 [152]3 years ago

8 0

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

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A car starts from rest and after 10 seconds is traveling at 20 m/s. Assuming that it continues to accelerate at the same rate it

Alenkinab [10]

Assuming that it continues to accelerate at the same rate it will take another 10 seconds to reach 40 m/s.

Answer:

Explanation:

Since the first question states that there is a change in the velocity from rest to 20 m/s in 10 seconds time interval. So the acceleration experienced by the car during this 10 seconds should be determined first as follows:

Acceleration = (final velocity-initial velocity)/Time

Acceleration = (20-0)/10 = 2 m/s².

So this means the car is traveling with an acceleration of 2 m/s².

As it is stated that the car continues to move with same acceleration, then in the second case, the acceleration is fixed as 2 m/s², initial velocity as 20 m/s and final velocity as 40 m/s. So the time taken for the car to reach this velocity with the constant acceleration value will be as follows:

Time = Change in velocity/Acceleration

Time = (40-20)/2 = 20/2=10 s

So again in another 10 seconds by the car to reach 40 m/s from 20 m/s. Similarly the car will take a total of 20 seconds to reach from rest to 40 m/s value for velocity.

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3 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75

katen-ka-za [31]

Answer:

Part A) 3899 kPa

Part B) 392.33 kJ/kg

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg

$\alpha$ $r_{1}$ = 621.2

The relative specific volume at state 2 is obtained from the compression ratio:

$\alpha$ $r_{2}$ = $\frac{\alpha r_{1} }{r}$

=621.2/ 8

= 77.65

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:

$P_{2}$ = $P_{1} r\frac{T_{2} }{T_{1} }$

= 95*8*673/300

= 1705 kPa

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

= 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):

$T_{3}$ = 1539 K

$\alpha r_{3}$ = 6.588

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

= 3899 kPa

<u>Part B</u>

The relative specific volume at state 4 is obtained from the compression ratio:

$\alpha r_{4}= r\alpha r_{3}$

= 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:

$T_{4}$ =775 K

$u_{4}$ = 571.74 kJ/kg

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$

<u>Part C </u>

The thermal efficiency is obtained from the work and the heat input:

η= $\frac{w}{q_{in} }$

=0.523

<u>Part D </u>

The mean effective pressure is determined from its standard relation:

MEP= $\frac{w}{\alpha_{1}- \alpha_{2} }$

= $\frac{w}{\alpha_{1}(1- \frac{1}{r} }$

= $\frac{rwP_{1} }{RT_{1} (r-1) }$

=495 kPa

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Answer:

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Explanation:

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$\omega=2\pi f$

where $\omega$ is the angular velocity, and $f$ is the frequency.

Frequency can also be represented as:

$f=\frac{1}{T}$

where $T$ is the period, (the time it takes to conclude a cycle)

with this, the angular velocity is:

$\omega=\frac{2\pi}{T}$

The period T of rotation around the sun 365 days, thus, the angular velocity:

$\omega=\frac{2\pi}{365days}=0.0172rad/day$

if we want the angular velocity in rad/second, we need to convert the 365 days to seconds:

Firt conveting to hous

$365days(\frac{24hours}{1day} )=8760hours$

then to minutes

$8760hours(\frac{60minutes}{1hour} )=525,600minutes$

and finally to seconds

$525,600minutes(\frac{60seconds}{1minute})=31,536x10^3seconds$

thus, angular velocity in rad/second is:

$\omega=\frac{2\pi}{31,536x10^3seconds}=1.99x10^{-7}rad/second$

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