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Fofino [41]
2 years ago
15

In a sample of 18-karat gold, 75 percent of the total mass is pure gold, while the rest is typically 16 percent silver and 9 per

cent copper. If the density of pure gold is rhogold=19.3g/cm3, while the densitites of silver and copper are respectively rhosilver=10.5g/cm3 and rhocopper=8.90g/cm3, what is the overall density rho18kt of this alloy of 18-karat gold?
Physics
1 answer:
ivann1987 [24]2 years ago
8 0

Answer:

15.57 g/cm³

Explanation:

\rho_g = Density of gold = 19.3 g/cm³

\rho_s = Density of silver = 10.5 g/cm³

\rho_c = Density of copper = 8.9 g/cm³

Assuming total mass as 1000 g

Volume of gold

V_g=\frac{0.75\times 1000}{19.3}\\\Rightarrow V_g=38.86\ cm^3

Volume of silver

V_g=\frac{0.16\times 1000}{10.5}\\\Rightarrow V_g=15.238\ cm^3

Volume of copper

V_c=\frac{0.09\times 1000}{8.9}\\\Rightarrow V_c=10.11\ cm^3

Density of the alloy would be

\rho=\frac{M}{V_g+V_s+V_c}\\\Rightarrow \rho=\frac{1000}{38.86+15.238+10.11}\\\Rightarrow \rho=15.57\ g/cm^3

The overall density of this alloy is 15.57 g/cm³

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Which of the following diagrams involves a virtual image ?
sergiy2304 [10]

Answer:

The third diagram

Explanation:

  • <u>A virtual image</u> is an image that can not be formed on a screen.
  • <u>A convex lens</u> can form both virtual and real image depending on the position of the object from the lens.
  • A virtual image in convex lens is formed when the object is placed between the focus and the optical center of the lens.
  • In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.
7 0
2 years ago
A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses
gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

4 0
2 years ago
A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
Korolek [52]

Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

3 0
2 years ago
Recall that weight is a force and is equal to m*g, where g is the acceleration due to gravity exerted by the Earth near the Eart
Marysya12 [62]

Answer:

145.43 N

Explanation:

Weight is given by (mg)

where m = mass of the body

g = acceleration due to gravity

mass is constant everywhere and is equal to 77.1 kg, both on the earth and on the moon.

But the acceleration due to gravity exerted by the moon near the moon's surface is 16.6% that of Earth,

g(moon) = 0.166 g(earth) = 0.166 × 9.8 = 1.6268 m/s²

Weight on the moon = mg(moon) = (77.1×1.6268) = 125.43 N

3 0
2 years ago
An apparatus is used to prepare an atomic beam by heating a collection of atoms to a temperature T and allowing the beam to emer
Zepler [3.9K]

Answer:

As shown in the attachment

Explanation:

The detailed steps and mathematical assumptions and manipulation is as shown in the attachment.

3 0
2 years ago
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