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liubo4ka [24]
3 years ago
9

A moving sidewalk has a velocity of 1.3 m/s north. If a man walks south on

Physics
1 answer:
drek231 [11]3 years ago
7 0
All of the above no cap
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If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th
Rufina [12.5K]
Based on internet sources, <span>the basic formulas are: v^2/r = (at)^2/r = a ==> at^2 = r ==> t = sqrt(r/a). 
</span>
<span>Assuming the missing units are mutually compatible, as in the following example, they don't need to be known. </span>
<span>Acceleration = 1.6 cramwells/s^2 </span>
<span>Radius = 150 cramwells </span>
<span>t = sqrt(150/1.6) = 9.68 s.

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8 0
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
3 years ago
Which factors affect the resistance of a material? Check all that apply.
shutvik [7]

Answer: Length, thickness, and temperature

Explanation:

I did it

6 0
3 years ago
You decide to roll a 0.11-kgkg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelen
Oksanka [162]

The de Broglie wavelength \lambda = 4.0\times 10^{-30}m

We know that

de Broglie wavelength = \lambda = \frac{h}{mv}\lambda = \frac{6.63\times 10^{-34}}{0.11\times 1.5 \times 10 ^{-3}}

\lambda = 4.0\times 10^{-30}m

<h3>What is de Broglie wavelength?</h3>

According to the de Broglie equation, matter can behave like waves, much like how light and radiation do, which are both waves and particles. A beam of electrons can be diffracted just like a beam of light, according to the equation. The de Broglie equation essentially clarifies the notion of matter having a wavelength.

Therefore, whether a particle is tiny or macroscopic, it will have a wavelength when examined.

The wave nature of matter can be seen or observed in the case of macroscopic objects.

To learn more about de Broglie wavelength with the given link

brainly.com/question/17295250

#SPJ4

3 0
1 year ago
Free point cuz i cewl like dat‍♀️‍♂️
DedPeter [7]
Yes thank u teehee




.................... x
8 0
2 years ago
Read 2 more answers
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