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Ad libitum [116K]

3 years ago

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A car traveling at 60 km/h will skid 30 m when its brakes are locked. If the same car is traveling at 180 km/h, what will be the

skidding distance? Be sure to show all work to support your answer.

1 answer:

cestrela7 [59]3 years ago

8 0

Converting 60 and 180 kmph into m/s
that's 16.67 m/s and 50 m/s
now
16.67*16.67 = 2*a*30
277.88/60 = a
4.631 m/s^2
as the vehicle is same it will decelerate with the same value
and thus
50*50 = 2*4.631*s
2500/9.262 = s
that's around 270 m

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A gas sample is heated from -20.0°C to 57.0°C and the volume is increased from 2.00 L to 4.50 L. If the initial pressure is 0.14

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Answer:

The answer is e.

Explanation:

We take:

$T_{1}=253K$

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$P_{1}=0.14atm$

$V_{2}=4.50 l$

$T_{2}=330K$

Taking the gas as an ideal gas, we can use the ideal gas law:

$\frac{PV}{nRT}$

⇒ $n=\frac{P_{1}V_{1}}{RT_{1}}$ ⇒ $n=0.013mol$

Then:

$P_{2}=\frac{nRT_{2}}{V_{2}}$ ⇒ $P_{2}=0.0811 atm$

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The answer is Data collected from many experiments could not be explained.

Explanation:

With pretty much every science theory, the theory can not change without valid experimentation by multiple professionals.

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Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

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The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

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substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

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The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

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Therefore, rays travelling parallel to the principle axis of a concave mirror will reflect out through the mirrors focus.

It is true.

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Explanation:

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