Ask question

Ask question

All categories

2 years ago

15

A sprinter reaches his maximum speed in 2.5sec from rest with constant acceleration. He then maintains that speed and finishes t

he 100m in the overall time of 9.60 seconds. Determine his maximum speed . 5. The cone falling with a speed strikes and penetrates the block of packing material. The acceleration of the cone after impact is = −

1 answer:

liubo4ka [24]2 years ago

3 0

The maximum Speed of the Sprinter from the velocity time graph of his motion is; 11.98 m/s

<h3>How to find the maximum speed?</h3>

We are given;

Initial Speed; u = 2.5 s

Total distance; d = 100 m

Total time; T = 9.6 s

The total distance is;

d = ¹/₂(9.6 + (9.6 - 2.5) * v

where v is maximum speed.

Thus;

¹/₂(9.6 + (9.6 - 2.5) * v = 100

16.7v = 200

v = 200/16.7

v = 11.98 m/s

Read more about Maximum Speed at; brainly.com/question/4931057

#SPJ1

You might be interested in

A body whose velocity is constant has a. positive acceleration b. negative acceleration g. zero acceleration d. all of the above

adoni [48]

Answer:

option (c) is the correct answer which is zero acceleration.

Explanation:

It is given in the question that the velocity is constant.

Now,

the options are provided in relation to the acceleration.

We know,

acceleration is rate of change of velocity per unit time i.e

acceleration = $\frac{dV}{dt}$

since, the change in velocity is given to be zero,

thus, dV/dt = 0

hence,

acceleration = 0

therefore, option (c) is the correct answer which is zero acceleration.

4 0

3 years ago

A wind turbine designer is considering installing a horizontal axis wind turbine at a location in Michigan. To increase the powe

bogdanovich [222]

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P = n * 1/2 *<em> p</em> *Av^3

where: n = turbine efficiency

<em>p = air density </em>

<em> </em>A = πd^2 / 4

v = speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

3 0

2 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 278C, and 75

Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process $T_{3}$ = 2042.56 K

(b). The value of pressure at the end of heat addition process $P_{3}$ = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle $E_{otto}$ = 0.4478

(d). The value of mean effective pressure of the cycle $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

Explanation:

Compression ratio $r_{p}$ = 8

Initial pressure $P_{1}$ = 95 k pa

Initial temperature $T_{1}$ = 278 °c = 551 K

Final pressure $P_{2}$ = 8 × $P_{1}$ = 8 × 95 = 760 k pa

Final temperature $T_{2}$ = $T_{1}$ × $r_{p} ^{\frac{\gamma - 1}{\gamma} }$

Final temperature $T_{2}$ = 551 × $8 ^{\frac{1.4 - 1}{1.4} }$

Final temperature $T_{2}$ = 998 K

Heat transferred at constant volume Q = 750 $\frac{KJ}{kg}$

(a). We know that Heat transferred at constant volume $Q_{S}$ = $m C_{v} ( T_{3} - T_{2} )$

⇒ 1 × 0.718 × ( $T_{3}$ - 998 ) = 750

⇒ $T_{3}$ = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

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

⇒ $P_{3}$ = $\frac{2042.56}{998}$ × 760

⇒ $P_{3}$ = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle $Q_{R} = m C_{v} ( T_{4} - T_{1} )$

For the compression and expansion process,

⇒ $\frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }$

⇒ $\frac{2042.56}{998} = \frac{T_{4} }{551}$

⇒ $T_{4}$ = 1127.7 K

Heat rejected $Q_{R}$ = 1 × 0.718 × ( 1127.7 - 551)

⇒ $Q_{R}$ = 414.07 $\frac{KJ}{kg}$

Net heat interaction from the cycle $Q_{net}$ = $Q_{S}$ - $Q_{R}$

Put the values of $Q_{S}$ & $Q_{R}$ we get,

⇒ $Q_{net}$ = 750 - 414.07

⇒ $Q_{net}$ = 335.93 $\frac{KJ}{kg}$

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ $Q_{net}$ = $W_{net}$

⇒ $W_{net}$ = 335.93 $\frac{KJ}{kg}$

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

$E_{otto} = 1- \frac{T_{1} }{T_{2} }$

Put the values of $T_{1}$ & $T_{2}$ in the above formula we get,

$E_{otto} = 1- \frac{551 }{998 }$

⇒ $E_{otto}$ = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure $P_{m}$ :-

We know that mean effective pressure of the Otto cycle is given by

$P_{m}$ = $\frac{W_{net} }{V_{s} }$ ---------- (1)

where $V_{s}$ is the swept volume.

$V_{s}$ = $V_{1} - V_{2}$ ---------- ( 2 )

From ideal gas equation $P_{1}$ $V_{1}$ = m × R × $T_{1}$

Put all the values in above formula we get,

⇒ 95 × $V_{1}$ = 1 × 0.287 × 551

⇒ $V_{1}$ = 0.6 $m^{3}$

From the same ideal gas equation

$P_{2}$ $V_{2}$ = m × R × $T_{2}$

⇒ 760 × $V_{2}$ = 1 × 0.287 × 998

⇒ $V_{2}$ = 0.377 $m^{3}$

Thus swept volume $V_{s}$ = 0.6 - 0.377

⇒ $V_{s}$ = 0.223 $m^{3}$

Thus from equation 1 the mean effective pressure

⇒ $P_{m}$ = $\frac{335.93}{0.223}$

⇒ $P_{m}$ = 1506.41 $\frac{k pa}{kg}$

This is the value of mean effective pressure of the cycle.

4 0

3 years ago

Which of the following behaviors should be avoided? Check all of the boxes that apply.

Ahat [919]

Answer:

c,d,e

Explanation:

rude, so you can keep safe and so you can really learn

7 0

2 years ago

Read 2 more answers

Mercury flows inside a copper tube 9m long with a 5.1сm inside diameter at an average velocity of 7.0 m/s. The inside surface te

svp [43]

Answer:

rate of heat transfer = 9085708.80 W

Explanation:

Given:

Inside diameter, D = 5.1 cm

= 5.1 x $10^{-2}$ m

Average velocity, V = 7 m/s

Mean temperature, T = (66+38) /2

= 52°C

Therefore kinematic viscosity at 52°C is ν = 0.104 X $10^{-6}$ $m^{2}$ / s

Prandtl no., Pr = 0.021

We know Renold No. is

Re = $\frac{V\times D}{\nu }$

Re = $\frac{7\times 5.1\times 10^{-2}}{0.104\times 10^{-6}}$

= 3.432 X $10^{6}$

Therefore the flow is turbulent.

Since the flow is turbulent and the ratio of L/D is greater than 60 we can use Dittua-Boelter equation.

Nu = 0.023 $Re^{0.8}$ . $Pr^{0.3}$

= 0.023 x $(3.432 \times10^{6})^{0.8}$ x $(0.021)^{0.3}$

= 1221.52

Since Nu = $\frac{h.D}{k}$

h = $\frac{k\times Nu}{D}$

= $\frac{9.4\times 1221.52}{5.1\times 10^{-2}}$

= 225143.3

Therefore rate of heat transfer, q = h.A(T- $T_{\infty }$

q= 225143.3 x 2πrh ( 66-38)

= 225143.3 X 2π X $\frac{5.1\times10^{-2}}{2}\times 9\times 28$

= 9085708.80 W

6 0

3 years ago