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Archy [21]
3 years ago
15

If you stood on Mars and lifted a 15kg pack,you would be pulling with a force greater than...

Physics
1 answer:
Oxana [17]3 years ago
6 0

Answer: See answers below.

Explanation: In this problem, we must be clear about the concept of weight. Weight is defined as the product of mass by gravitational acceleration.

We must be clear that the mass is always preserved, that is, the mass of 15 [kg] will always be the same regardless of the planet where they are.

where:

W = weight [N] (units of Newtons)

m = mass = 15 [kg]

g = gravity acceleration [m/s²]

Since we have 9 places with different gravitational acceleration, then we calculate the weight in each of these nine places.

Mercury

Venus

Moon

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

In this problem, we must be clear about the concept of weight. Weight is defined as the product of mass by gravitational acceleration.

We must be clear that the mass is always preserved, that is, the mass of 15 [kg] will always be the same regardless of the planet where they are.

where:

W = weight [N] (units of Newtons)

m = mass = 15 [kg]

g = gravity acceleration [m/s²]

Since we have 9 places with different gravitational acceleration, then we calculate the weight in each of these nine places.

Mercury

Venus

Moon

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

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At what net rate does heat radiate from a 300-m^2 black roof on a night when the roof's temperature is 33.0°C and the surroundin
Fynjy0 [20]

Answer:

24445.85 J/s

Explanation:

Area, A = 300 m^2

T = 33° C = 33 + 273 = 306 k

To = 18° C = 18 + 273 = 291 k

emissivity, e = 0.9

Use the Stefan's Boltzman law

E = \sigma  \times e \times A\times\left ( T^4 -T_{0}^{4}\right )

Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.

The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4

By substituting the values

E = 5.64 \times 10^{-8}\times 0.9 \times 300 \times  (306^{4}-291^{4})

E = 24445.85 J/s

7 0
4 years ago
The resolution of a camera or other optical system is determined by the relationship between what two scales?
devlian [24]

Answer:

d.The wavelength of light and the size of the aperture

Explanation:

<em>The resolution power of an optical system is the smallest distance between two points that the device can distinguish clearly.</em>

It has the following relationship:

r=\frac{\lambda}{2n}

where:

r = minimum resolvable distance

n = numerical aperture

\lambda= wavelength of the light used for viewing

From above mathematical equation it is clear that:

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(Note, that smaller the value of "r" the more finer details of the image visible through the device.)

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3 years ago
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4 years ago
The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental
pav-90 [236]

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

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Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

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4 0
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