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3 years ago

How can we predict how high a roller coaster can go knowing its velocity?

1 answer:

6 0

“to predict the speed that a coaster will reach before it is ever placed on the track. ... When coaster goes up by height h then its gravitational potential energy increases by the amount m.g.h where m = mass of coaster, h = height of coaster and g = gravitational acceleration due to Earth”

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List each of the parts of the electromagnetic spectrum in order of lowest to highest frequency. Write down one use for each one,

Juliette [100K]

Answer:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves.

6 0

3 years ago

The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0

2 years ago

Jack has a weight of 300 N and sits 2.0 m from the pivot of see - saw. Jill has a weight of 450 N and sits 1.5 m from pivot. Who

coldgirl [10]

Answer:

Jill will move down first

Explanation:

3 0

3 years ago

What is the frequency of a mechanical wave that has a

kakasveta [241]

Explanation:

v= (f) x ( lambda)

1.7 ms^-1/12.05 m = f =o.14 hz

5 0

3 years ago

A 60-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and that the

Mariulka [41]

Answer:

part (a). 176580 J

part (b). 197381 J

Explanation:

Given,

Density of the chain = $\rho\ =\ 10\ kg/m.$
Length of the chain = L = 60 m
Acceleration due to gravity = g = 9.81 $m/s^2$

part (a)

Let dy be the small element of the chain at a distance of 'y' from the ground.

mass of the small element of the chain = $\rho dy$

Work done due to the small element,

$dw\ =\ \rho g (60\ -\ y)dy\\$

Total work done to wind the entire chain = w

$w\ =\ \displaystyle\int_{0}^{L} \rho g(60\ -\ y)dy\\\Rightarrow w\ =\ \rho g\left |(60y\ -\ \dfrac{y^2}{2})\ \right |_{0}^{60}\\\Rightarrow w\ =\ 10\times 9.81\times (60\times 60\ -\ \dfrac{60^2}{2})\\\Rightarrow w\ =\ 176580\ J$

part (b)

mass of the block connected to the chain = m = 35 kg

Total work done to wind the chain = work done due to the chain + work done due to the mass

$\therefore W\ =\ w\ +\ mgL\\\Rightarrow W\ =\ 176580\ +\ 35\times 9.81\times 60\\\Rightarrow W\ =\ 176580\ +\ 20601\\\Rightarrow W\ =\ 197381\ J$

4 0

2 years ago