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3 years ago

When a crown of mass 14.7 kg is submerged in water, an accurate scale reads only 13.4 kg. What material is the crown made of?

Physics

1 answer:

Bogdan [553]3 years ago

3 0

Answer:

ρ = 1.13 10⁴ km/m³

Explanation:

For this exercise we use Newton's equilibrium equation

B –W + W_scale = 0

Where B is the thrust and W_scale is the balance reading

The push is given by Archimedes' law

B = ρ_water g V

B = W- W_scale

B = m g - m_scale g

Let's calculate

B = 14.7 9.8 - 13.4 9.8

B = 12.74 N

ρ_water g V = 12.74

V = 12.74 / ρ_water g

V = 12.74 / 1000 9.8

V = 0.0013 m³

Let's use density

ρ = m / V

We replace

ρ = 14.7 / 0.0013

ρ = 1.13 10⁴ km/m³

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3. How does the mass of a proton compare to the mass of an electron?

o-na [289]

Answer:

Proton, stable subatomic particle that has a positive charge equal in magnitude to a unit of electron charge and a rest mass of 1.67262 × 10−27 kg, which is 1,836 times the mass of an electron.

7 0

2 years ago

When you go for a walk which of the following forces is paired with the force of friction on your shoe

Ann [662]

Answer:

static

Explanation:

static friction pushes in the direction you are walking.

7 0

3 years ago

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4 ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a) final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5 ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a) final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀ ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a) final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

3 0

2 years ago

I don’t have a question because I posted a photoz

Scorpion4ik [409]

Part a:

$Q_{1}$ = 56

$Q_{2}$ = 60

$Q_{3}$ = 63

The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The $Q_{2}$ is the overall medium, $Q_{1}$ is the medium of the first half, and $Q_{3}$ is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between $Q_{1}$ and $Q_{3}$ . In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + $\frac{7}{2}$ ) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0

1 year ago

Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C

DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0

3 years ago