These type of winds are located between 30 and 60 degrees latitude on the globe.

Explain in terms of energy flow how a cold pack works on a sprained ankle

sergij07 [2.7K]

The <span>flow of how a cold pack works on a sprained ankle is based on the second law of thermodynamics which states that energy will flow from a higher to a lower temperature. So your body heat will flow to the cold pack in which you will feel the coldness of the pack.</span>

7 0

2 years ago

Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po

gregori [183]

Answer:

E = 0 r <R₁

Explanation:

If we use Gauss's law

Ф = ∫ E. dA = $q_{int}$ / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

E = 0 r <R₁

6 0

2 years ago

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, $d=24.44\times 2=48.88\ m$

(a) Acceleration, $a=-4\ m/s^2$

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -4}$

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, $a=-8\ m/s^2$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -8}$

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0

3 years ago

the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its

GalinKa [24]

Answer:

please find attached pdf

Explanation:

Download pdf

8 0

2 years ago

A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol) at atmospheric pres

Vikentia [17]

Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

V = L^3 = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

P*V = n*R*T

n = P*V / R*T

n = 101.3*10^3 * 0.001 / 8.314*300

n = 0.04061 moles

- Compute the mass of the gas:

m = n*M

m = 0.04061*28.9

m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

F_g = m*g

F_g = 1.174*9.81*10^-3

F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

F_c = P*A

F_c = (101.3*10^3)*(0.1)^2

F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

4 0

2 years ago