when we find the distance we will add all the blocks so
distance = 6+6+4
distance = 14blocks
when we find the displacement we will add and minus too
As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = <em><u>4blocks</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>west</u></em>
Answer:3.51
Explanation:
Given
Coefficient of Friction 
Consider a small element at an angle \theta having an angle of 
Normal Force

Friction 

and 







There is not enough information to draw a conclusion about
Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
Answer
-Directly; outside air pressure
Vapor pressure is directly related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the outside air pressure.
Explanation;
-As the temperature of a system increases, the average kinetic energy of the molecules increases in both the liquid and gas phases.
-A higher average kinetic energy facilitates the escape of molecules from the liquid phase into the gas phase. At the same time, the rate of return of gas phase molecules to the liquid also increases. A new equilibrium point is reached at a higher gaseous vapor pressure. The increase in vapor pressure with temperature is exponential.