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3 years ago

13

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,

the acceleration due to gravity is what factor times the value of g at the Earth's surface?

1 answer:

Taya2010 [7]3 years ago

7 0

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

$F=\frac{GM_1M_2}{r^2}$

Force $F=mg'=\frac{GMm}{4R^2}-----1$

Force on earth surface $F=mg=\frac{GMm}{R^2}------2$

Divide 1 and 2 we get

$\frac{g'}{g}=\frac{R^2}{4R^2}$

$g'=\frac{g}{4}$

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Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

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3 years ago

Two concrete spans of a 380 m long bridge are

Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

$L_i = \frac{L}{2}=\frac{380}{2}=190 m$

Due to the increase in temperature, the length of each span increases according to:

$L_f = L_i(1+ \alpha \Delta T)$

where

$L_i = 190 m$ is the initial length of one span

$\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1}$ is the temperature coefficient of thermal expansion

$\Delta T=20^{\circ}C$ is the increase in temperature

Substituting,

$L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m$

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

$h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m$

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3 years ago

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ololo11 [35]

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3 years ago

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Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th

Anit [1.1K]

Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components. While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

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3 years ago

Please awnser and show the ways

topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0

2 years ago