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2 years ago

What is the density of a sample of ore that has a mass of 74.0 g and occupies 20.3 cm3?

1 answer:

4 0

Density =mass/ volume
Data:
mass=74 g
volume=20.3 cm³

Density=74 g / 20.3 cm³≈3.65 g/cm³

Answer: The density of ore is 3.65 g/cm³ (the ore could be "pyrope")

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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m

0.80 m

1.25 m

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0

2 years ago

Read 2 more answers

A future use of space stations may be to provide hospitals for severely burned persons. it is very painful for a badly burned pe

natta225 [31]

1.5 minutes per rotation. The formula for centripetal force is A = v^2/r where A = acceleration v = velocity r = radius So let's substitute the known values and solve for v. So F = v^2/r 0.98 m/s^2 = v^2/200 m 196 m^2/s^2 = v^2 14 m/s = v So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin. Its circumference is 2*pi*r, so C = 2 * 3.14159 * 200 m C = 1256.636 m And we need a velocity of 14 m/s, so 1256.636 m / 14 m/s = 89.75971429 s Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.

5 0

2 years ago

A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o

kvasek [131]

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
x = 500 nm / 5.96
x = 83.90 nm

6 0

2 years ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Lisa [10]

On comparing values , we see that student which has the largest percent error is A. Student 4: 9.61 m/s2 .

Explanation:

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

A. Student 4: 9.61 m/s2

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 1.738%$ %.