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3 years ago

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What are two examples of goods and two examples of services

1 answer:

julsineya [31]3 years ago

8 0

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

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A 1.3-kg ball is attached to the end of a 0.8-m string to form a pendulum. This pendulum is released from rest with the string h

Natali5045456 [20]

Answer:

2.9 m/s

Explanation:

Momentum will be conserved

Speed of the ball just before collision is

v = √2gh = √(2(9.8)(0.8)) = 3.96 m/s

The initial momentum is 1.3(3.96) = 5.15 kg•m/s

The block takes away momentum of 0.6(2.2) = 1.32 kg•m/s

Leaving the ball with momentum of 5.15 - 1.32 = 3.83 kg•m/s

vf(ball) = 3.83 / 1.3 = 2.946... ≈ 2.9 m/s

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3 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

frez [133]

Answer:

<em>1.49 x </em> $10^{11}$ <em></em>

<em></em>

Explanation:

Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>

Mathematically, this can be stated as

$T^{2}$ ∝ $R^{3}$

<em>to remove the proportionality sign we introduce a constant</em>

$T^{2}$ = k $R^{3}$

k = $\frac{T^{2} }{R^{3} }$

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x $10^{11}$

for mars, constant k will be

k = $\frac{687^{2} }{(2.279*10^{11}) ^{3} }$ = 3.987 x $10^{-29}$

For Earth, orbital period T is 365 days, therefore

$365^{2}$ = 3.987 x $10^{-29}$ x $R^{3}$

$R^{3}$ = 3.34 x $10^{33}$

R =<em> 1.49 x </em> $10^{11}$ <em></em>

8 0

3 years ago

Read 2 more answers

At what constant rate of acceleration will a car starting from rest cover 18m in the first 3 seconds?

irga5000 [103]

69 hehehehe hehehehe here

7 0

3 years ago

It took a bulldozer 62,000 J of work to move a rock 30 m. It took 5 minutes. How much force did the bulldozer have to apply?

NeX [460]

Answer:

A (2066,6 N)

Explanation:

Use the Work formula

62.000J = F . 30

62.000/30 = 2066,6 N

The amout of time it took to move the rock doesn´t matter at all.

It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.

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Could someone please help me with this problem? I’m thinking that they both will have the same force because they have the same

Simora [160]

You would be correct

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3 years ago