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3 years ago

Sound travels fastest through A) space. B) cool air. C) warm air. D) a metal spoon.

Physics

2 answers:

Mademuasel [1]3 years ago

7 0

A metal spoon. Because metal travels the fastest

steposvetlana [31]3 years ago

4 0

I took the test the answer is Metal Spoon because sound always travels fastest through solids!

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It takes 15 minutes to walk to the mall from your house at an average velocity of 1.1 m/s to the east. What is the displacement

Anna11 [10]

15 minutes = 15*60 = 900 seconds

900x1.1 = 990 m

4 0

3 years ago

NEEDD SOME HELP ASAP HELP MEHHH 50 pOINTS

Vadim26 [7]

An electron shell can hold 2(n^2) electrons (technically) where n is the shell number, i.e. shell 1 can hold 2, shell 2 can hold 8, 3 holds 18 and so on.
The atomic number of Nitrogen is 7, i.e. it has 7 electrons (to match its 7 protons, assuming it isn't an ion).

With the atomic number, you simply start from shell 1 and work out. So we put 2 electrons in shell 1, leaving us with 5 left. Shell 2 can hold 6 so we can fit all 5 in.

In other words, you should have 2 electron shells on the atom, shell 1 with 2 e- and shell 2 with 5 e-.

3 0

3 years ago

Read 2 more answers

A hydrometer is made of a tube of diameter 2.3cm.The mass of the tube and it's content is 80g. If it floats in a liquid density

iris [78.8K]

Answer:

The depth to which the hydrometer sinks is approximately 24.07 cm

Explanation:

The given parameters are;

The diameter of the hydrometer tube, d = 2.3 cm

The mass of the content of the tube, m = 80 g

The density of the liquid in which the tube floats, ρ = 800 kg/m³

By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid

When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer

Therefore;

The weight of the liquid displaced = The weight of the hydrometer, W = m·g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ W = 80 g × g

The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ

V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³

The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged, $V_h$

$V_h$ = A × h

Where;

A = The cross-sectional area of the tube = π·d²/4

h = The depth to which the hydrometer sinks

h = $V_h$ /A

∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm

The depth to which the tube sinks, h ≈ 24.07 cm.

3 0

3 years ago

A model air rocket with a mass of 50.g is free to travel along a horizontal track. It begins from rest. After 2.0s, the rocket h

katen-ka-za [31]

Answer:

$v_r=5.89\ m.s^{-1}$

Explanation:

Given:

mass of rocket, $m_r=50\ g$

time of observation, $t=2\ s$

mass lost by the rocket by expulsion of air, $m_a=10\%\ of m_r=5\ g$

velocity of air, $v_a=53\ m.s^{-1}$

Now the momentum of air will be equal to the momentum of rocket in the opposite direction: (Using the theory of elastic collision)

$m_a.v_a=(m_r-m_a)\times v_r$

$5\times 53=(50-5)\times v_r$

$v_r=5.89\ m.s^{-1}$

7 0

3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago