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1 year ago

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An engineer wants to design an oval racetrack such that 3.20 × 10 3 lb racecars can round the exactly 1000 ft radius turns at 10

2 mi/h without the aid of friction. She estimates that the cars will round the turns at a maximum of 175 mi/h. Find the banking angle θ necessary for the race cars to navigate the turns at 102 mi/h without the aid of friction.

1 answer:

Reptile [31]1 year ago

4 0

Answer:

The banking angle necessary for the race cars is 34.84°

Explanation:

For normal reaction the expression is:

$\\Nsin\theta = \frac{mv^{2} }{R} =Fc\\tan\theta =\frac{v^{2} }{Rg} \\\theta =tan^{-1} (\frac{v^{2} }{Rg} )\\\theta =tan^{-1} (\frac{(102*0.447)^{2} }{1000*0.3048*9.8} )=34.84$

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Which formula is used to find fluctuation of the shape of body

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Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

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2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

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2 years ago

The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

$\frac{0.60}{R_A^2}=\frac{1}{R_B^2}$

So the ratio between the radii of the two planets is

$\frac{R_A}{R_B}=\sqrt{0.60}=0.775$

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2 years ago

Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate

Wittaler [7]

Answer:

455165.278 m

Explanation:

P = Power = 3.7 W

v = Velocity = 10.7 m/s

Amount of fat = 4 g

1 gram of fat provides about 9.40 (food) Calories

Energy given by 4 g of fat

$E=4\times 9.4\times 4186\\\Rightarrow E=157393.6\ J$

Time required to burn the fat

$t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s$

Distance traveled by the bird

$s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m$

The bird will fly 455165.278 m

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2 years ago

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The trough of the sine curve used to represent a sound wave corresponds to

iren [92.7K]

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

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