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3 years ago

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A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×

105 N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?

1 answer:

otez555 [7]3 years ago

4 0

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, $m=1.01\times 10^7\ kg$

Force applied on the tracks, $F=7.5\times 10^5\ N$

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

$F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s$

So, the required time is 300.3 seconds.

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Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

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3 years ago

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

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3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

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3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

a)

$v = 4.048 *10^6$ m/s

b)

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

a)

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

b)

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

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2 years ago

Can a solution be separated by filtration?

alexgriva [62]

Answer:

Filtration is a method for separating an insoluble solid from a liquid. When a mixture of sand and water is filtered: the sand stays behind in the filter paper (it becomes the residue ) the water passes through the filter paper (it becomes the filtrate )

Explanation:

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