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The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

8 0

3 years ago

These are the most widely used tools and most often abuse tool

Mars2501 [29]

Where is the picture in this problem? How am I supposed to answer if I can’t see any footage taken from this problem.

4 0

2 years ago

Read 2 more answers

Why is it reasonable to say that no system is 100% efficient?

Virty [35]

Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.

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2 years ago

6) Describe the differences between the troposphere and stratosphere.

vodka [1.7K]

Answer:

frejfru8758u37

Explanation:

i need points

8 0

2 years ago

Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient press

lutik1710 [3]

Solution :

Methods for selling pressure of a distillation column :

a). Set, $\text{based on the pressure required to condensed}$ the overhead stream using cooling water.

(minimum of approximate 45°C condenser temperature)

b). Set, $\text{based on highest temperature}$ of bottom product that avoids decomposition or reaction.

c). Set, $\text{based on available highest }$ not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

a). The rate of reaction is taster. This results in a small reactor or high phase conversion.

b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.

Run the reaction at an evaluated pressure because :

The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.

The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.

7 0

2 years ago