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3 years ago

PLEASE HURRY! The highest speed ever achieved on a bicycle was reached by John Howard of the United

Physics

2 answers:

Yuki888 [10]3 years ago

4 0

Answer:

Say, the speed of the bicycle before it was decelerated is 245 km/h. Then, John would like to decrease his bicycle's speed by 20%. That means, after the deceleration, he still has 80% of the initial speed.

Let's find out how fast is 80% of the initial speed.

80% * 245 km/h = 196 km/h

So let's say, the terminal speed of the bicycle (after deceleration) is 196 km/h.

Find the difference between the initial speed and terminal speed:

245 km/h - 196 km/h = 49 km/h

Now change the unit of km/h to m/s:

49 km/h = 49 / 3.6 m/s = 13.61 m/s

now determine the time needed to decelerate 20% of the initial speed.

time (t) = difference of velocity (ΔV) / deceleration (d)

t = 13.61 m/s / -3.00 m/s^2

t = 4.537 seconds

Explanation:

smart

pashok25 [27]3 years ago

3 0

Well, to begin with, your first number gang somewhat aglay. The land speed record that John Howard set on his bicycle in 1985 was 152.2 miles per hour, which works out to 68.04 m/s. So I can see where you got the 6 and the 8 from, but your little decimal point snuck over one place when you weren't looking.

I'll use your number to answer the question. If my solution turns out to be wrong, then it's because you copied the number wrong, and you'll have to work it out again with an initial speed of 68.1m/s.

Initial speed = 6.81 m/s

Final speed = 5.44 m/s

Amount of slowing down = 1.37 m/s

Rate at which the brakes slow you down = 3 m/s each second

Time needed to slow down 1.37 m/s = (1.37 m/s) / (3 m/s^2)

That's 0.457 second. (obviously absurd)

If initial speed = 68.1 m/s

Then amount of slowing down = 62.66 m/s

Time needed at -3 m/s^2 = (62.66/3)

That's 20.9 seconds. Much more reasonable.

By the way, John Howard's record was broken 10 yrs later, in 1995 .

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4 years ago

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

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3 years ago

Renald completed only one trial of his experiment. what effect will this most likely have?

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3 years ago

Find the pressure necessary to change a volume of steel by

kramer

Answer:

160×10⁶ bar

Explanation:

Bulk modulus is defined as:

K = -V ΔP / ΔV

where V is the initial volume,

ΔP is the change in pressure,

and ΔV is the change in volume.

Given:

K = 160×10⁹ Pa

ΔV = -0.01V

160×10⁹ Pa = -V ΔP / (-0.01V)

160×10⁹ Pa = ΔP / 0.01

ΔP = 160×10¹¹ Pa

ΔP = 160×10⁶ bar

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