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4 years ago

Which molecules would most likely cause a liquid to have the lowest viscosity? large, polar molecules small, nonpolar molecules

Physics

2 answers:

Debora [2.8K]4 years ago

5 0

Smaller nonpolar molecules

Ede4ka [16]4 years ago

4 0

Answer:

Small, non polar molecules.

Explanation:

Viscosity is a property of a liquid which is similar to friction or drag. When a liquid has high viscosity, it flows slowly and when it flows easily, the liquid is said to be less viscous.

Polar molecules will share the electrons in the covalent bonds unequally. Liquids that have polar bonds are said to have higher viscosity.

In non polar bonds, atoms have equal electronegativity.

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A mass is oscillating horizontally on a spring. At the locations A, B, C, D, and E, photogates are used to measure the speed of

svetoff [14.1K]

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is $PE_{elastic} = 50J$

The kinetic energy at point D is $KE = 75J$

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is $E_ {m} = 100J$

So at point B

$E_{m} = PE_{elastic} +KE$

$100 = PE_{elastic} + KE$

KE at point B is 50J

So $PE_{elastic} = 100 - 50 = 50J$

Now at point D

$E_{m} = PE_{elastic} +KE$

$PE_{elastic}$ at point D is 25J

So $KE = 100 - 25 = 75J$

4 0

3 years ago

What is the molecular formula for the hydrocarbon in the mass spectrum above?

PtichkaEL [24]

What is the structure of a hydrocarbon that has $\mathrm{M}^{+}=120$ in its mass spectrum and has the following $1 \mathrm{H}$ NMR spectrum? 7.25 $\delta(5 \mathrm{H}, \text { broad singlet); } 2.90 \delta(1 \mathrm{H}, \text { septet, } J=7 \mathrm{Hz}) ; 1.22 \delta(6 \mathrm{H},\text { doublet, }$ $J=7 \mathrm{Hz})$

6 0

3 years ago

In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of

Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, $a_{Net}$ = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - $a_{Net}$ × t

∴ t = (v₁ - v₂)/ $a_{Net}$ = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2· $a_{Net}$ ·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0

3 years ago

Yung hangs a mass from a rod, sets the pendulum in simple harmonic motion, and determines the period of the pendulum. How could

kow [346]

He could use a shorter rod.

8 0

3 years ago

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time accordin

Doss [256]

Answer:

$t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}$

Explanation:

After time "t" the angular position of A is given as

$\theta_a = \theta_o + \omega_o t + \frac{1}{2}\alpha t^2$

now we know that B start motion after time t1

so its angular position is also same as that of position of A after same time "t"

so we have

$\theta_b = \theta_o + \frac{\omega_o}{2} (t - t_1) + \frac{1}{2}(2\alpha) (t - t_1)^2$

now since both positions are same

$\theta_a = \theta_b$

$\omega_o t + \frac{1}{2}\alpha t^2 = \frac{\omega_o}{2}(t - t_1} + \alpha(t - t_1)^2$

$2\omega_o t + \alpha t^2 = \omega_o(t - t_1) + 2\alpha(t - t_1)^2$

$t - t_1 = \frac{-\omega_o + \sqrt{\omega_o^2 + 8\alpha(2\omega_o t + \alpha t^2)}}{4\alpha}$

3 0

4 years ago