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ollegr [7]
3 years ago
7

Can an element be a molecule

Physics
2 answers:
irinina [24]3 years ago
8 0
Yes molecules are the smallest particle and elments are usually made up of molecules
Serga [27]3 years ago
6 0

an element can make a molecule. so technically yes.

You might be interested in
A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before sto
ladessa [460]

Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

4 0
2 years ago
At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−
vredina [299]

Answer:

a)  Δx = 49.23 mi , b)  Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

     a = dv / dt

     ∫dv = ∫ a dt

we integrate and evaluators

      v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

       1+ 8t = u

       8 dt = du

       v - v₀ = -1280 ∫ u⁻³ du / 8

       v -v₀ = -1280 / 8 (-u⁻²/2)

       v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

      v- 80 = 80 [(1 + 8t)⁻² - 1]

      v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

     v = dx / dt

    dx = vdt

    x-x₀ = 80 ∫ (1-8t)⁻² dt

    x-x₀ = 80 ∫ u⁻² dt / 8

    x-x₀ = 80 (-1 / u)

    x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

   x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

     x = x₀

t = 0.2 h

    x-x₀ = -80 [1 / (1 +8 02) -1]

    x-x₀ = 49.23

displacement is

  Δx = x (0.2) - x (0)

   Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

     x- x₀ = -80 [1 / (1+ 8 0.4) -1]

     x-x₀ = 55 mi

    Δx = x (0.4) - x (0.2)

     Δx = 55 - 49.23

     Δx = 5.77 mi

3 0
3 years ago
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature
Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0
3 years ago
Dark matter may explain _____.
Lisa [10]
Dark matter may explain <span>unexpected orbital velocities of stars in galaxies.</span>
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The answer to this question is C
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