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3 years ago

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Which contribution did Kepler make to viewing the solar system?

1 answer:

amm18123 years ago

8 0

He proposed the sun-centered model of the solar system.

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Which of the following does NOT create a mechanical wave? Text to speech

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Light coming through a window!!

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3 years ago

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It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibilty can be described by Hooke

Varvara68 [4.7K]

<span>k = 1.7 x 10^5 kg/s^2 Player mass = 69 kg Hooke's law states F = kX where F = Force k = spring constant X = deflection So let's solve for k, the substitute the known values and calculate. Don't forget the local gravitational acceleration. F = kX F/X = k 115 kg* 9.8 m/s^2 / 0.65 cm = 115 kg* 9.8 m/s^2 / 0.0065 m = 1127 kg*m/s^2 / 0.0065 m = 173384.6154 kg/s^2 Rounding to 2 significant figures gives 1.7 x 10^5 kg/s^2 Since Hooke's law is a linear relationship, we could either use the calculated value of the spring constant along with the local gravitational acceleration, or we can simply take advantage of the ratio. The ratio will be both easier and more accurate. So X/0.39 cm = 115 kg/0.65 cm X = 44.85 kg/0.65 X = 69 kg The player masses 69 kg.</span>

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3 years ago

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Which phrase describes an irregular galaxy?

viktelen [127]

Answer:b

Explanation:

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3 years ago

Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Assume the mower moves wit

bekas [8.4K]

Answer:

t = 11.9 h = 0.49 day

Explanation:

First of all, we will find the area covered by the mower per unit time:

$Rate\ of\ Area\ Covered\ by\ Mower = vw$

where,

v = speed of mower = 1 km/h = 1000 m/h

w = width of mower = 0.6 m

Therefore,

$Rate\ of\ Area\ Covered\ by\ Mower = (1000\ m/h)(0.6\ m)\\Rate\ of\ Area\ Covered\ by\ Mower = 600\ m^{2}/h$

Now, the standard football ground has an area of:

A = 105 m x 68 m = 7140 m²

Therefore, The time required to mow the lawn will be:

$Time\ to\ Mow\ the\ Lawn = t = \frac{A}{Rate\ of\ Area\ Covered\ by\ Mower}\\t = \frac{7140\ m^{2}}{600\ m^{2}/h}$

<u>t = 11.9 h = 0.49 day</u>

8 0

3 years ago

X-rays with an energy of 301 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 77.5^{\circ}

Alex73 [517]

Answer:

$6.03\cdot 10^{-12} m$

Explanation:

First of all, we need to find the initial wavelength of the photon.

We know that its energy is

$E=301 keV = 4.82\cdot 10^{-14}J$

So its wavelength is given by:

$\lambda = \frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.82\cdot 10^{-14} J}=4.13\cdot 10^{-12}m$

The formula for the Compton scattering is:

$\lambda' = \lambda +\frac{h}{mc}(1-cos \theta)$

where

$\lambda$ is the original wavelength

h is the Planck constant

m is the electron mass

c is the speed of light

$\theta=77.5^{\circ}$ is the angle of the scattered photon

Substituting, we find

$\lambda' = 4.13\cdot 10^{-12} m +\frac{6.63\cdot 10^{-34} Js)}{(9.11\cdot 10^{-31}kg)(3\cdot 10^8 m/s)}(1-cos 77.5^{\circ})=6.03\cdot 10^{-12} m$

7 0

3 years ago