Based on the path of a light ray through air, water, and glass, which medium has the greatest index of refraction?

What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a

Vikki [24]

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure = P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

5 0

3 years ago

A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in

Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J

But ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0

2 years ago

Read 2 more answers

A drag racing vehicle travels from zero to a hundred miles per hour and 5 seconds north what is acceleration

avanturin [10]

Acceleration(a) is the change in velocity(Δv) over the change in time(Δt). so just divide your velocity and time.

6 0

2 years ago

A sample of helium behaves as an ideal gas as energy is

alisha [4.7K]

Answer:

0.0321 g

Explanation:

Let helium specific heat $c_h = 5.193 J/g K$

Assuming no energy is lost in the process, by the law of energy conservation we can state that the 20J work done is from the heat transfer to heat it up from 273K to 393K, which is a difference of ΔT = 393 - 273 = 120 K. We have the following heat transfer equation:

$E_h = m_hc_h \Delta T = 20 J$

where $m_h$ is the mass of helium, which we are looking for:

$m_h = \frac{20}{c_h \Delta T} = \frac{20}{5.193 * 120} \approx 0.0321 g$

4 0

3 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

3 years ago