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3 years ago

5

he acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. The

weight of a space probe on the surface of Mars is about _____.

1 answer:

Ahat [919]3 years ago

5 0

1/3 the weight than it is on earth

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What are the main activities involved in studying physics?

Firlakuza [10]

The main activity that is involved in studying of physics is the study of natural laws. The study of physics has to do with many aspects of the universe. Physics majorly looks into the natural laws that operate in the universe and describe how they affect matter in relation to time.

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3 years ago

A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

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3 years ago

The infant's tendency to turn its head toward things that touch its cheek is known as the

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I think it is <em><u>Rooting</u></em><em> </em><u><em>Reflex</em></u>

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3 years ago

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You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$

Using this Area, we find the diameter of the wire:

$D = \sqrt{\frac{4A}{\pi}}$

$D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}$

$D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm$

To find the length, we multiply the two equations stated initially:

$mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}$

$l^2 = 8.534\\l = 2.92 \ m$

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2 years ago

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