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Sergeeva-Olga [200]
3 years ago
13

You drive on Interstate 10 from San Antonio to Houston, half the time at 54 km/h and the other half at 118 km/h. On the way back

you travel half the distance at 54 km/h and the other half at 118 km/h. What is your average speed?
Physics
2 answers:
Irina-Kira [14]3 years ago
7 0

Answer:

Average speed: 86 km/h

Explanation:

Driving from San Antonio to Houston:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

Driving way back:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

As in both routes we have the same average speed, then the average speed for the whole trip is 86 km/h

pogonyaev3 years ago
7 0

Answer:

v_{avg} = 79.7 km/h

Explanation:

As we move from San Antonio to Houston

let the distance is "d" from Antonio to Houston

Half the time it moves with 54 km/h and next half the time it moves 118 km/h

so we will have

54 T + 118 T = d

T = \frac{d}{172}

so total time is

2T = \frac{d}{86} = 0.0116 d

now while his return journey half the distance he move with 54 km/h and next half distance with speed 118 km/h

so we have time of return journey

T' = \frac{d/2}{54} + \frac{d/2}{118}

so now we have

T' = 0.0135 d

now for the average speed we know that

v_{avg} = \frac{distance}{time}

v_{avg} = \frac{d + d}{0.0116 d + 0.0135 d}

v_{avg} = 79.7 km/h

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Please awnser and show the ways​
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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

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<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

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<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

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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

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