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Natasha_Volkova [10]

2 years ago

11

A 1840-kg car travels on a banked, horizontal curve of diameter 225 m. Find the maximum safe speed if the coefficient of frictio

n between the tires and the road is 0.90 and the banking angle is 6.0°. Additionally, what net force does the car experience in this case?

1 answer:

BigorU [14]2 years ago

8 0

Answer:

35 m/s

Explanation:

m = 1840 kg

2R = 225 m

R = 112.5 m

μ = 0.9

θ = 6 degree

use the formula for the banking of road

$v = \sqrt{\frac{Rg(Sin\theta +\mu Cos\theta )}{Cos\theta -\mu Sin\theta }}$

$v = \sqrt{\frac{112.5\times 9.8(Sin6 +0.9 Cos6 )}{Cos6 -0.9Sin6 }}$

v = 35 m/s

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The equation that defines the linear moment is given by

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$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

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We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

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B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

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an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
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$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
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Product of extremes equals product of means:
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$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

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Answer:

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Explanation:

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