Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Answer:
Explanation:
Given
Ship A velocity is 40 mph and is traveling 35 west of north
Therefore in 2 hours it will travel 
thus its position vector after two hours is

similarly B travels with 20 mph and in 2 hours
![=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}](https://tex.z-dn.net/?f=%3D20%5Ctimes%202%3D40%20miles%20%3C%2Fp%3E%3Cp%3EIts%20position%20vector%5Btex%5Dr_B%3D40sin80%5Chat%7Bi%7D%2B40cos80%5Chat%7Bj%7D)
Thus distance between A and B is



Velocity of A

Velocity of B

Velocity of A w.r.t B


Answer:
Correct, is there another part to the question?
Answer:
Water normally freezes at 0°C (32°F). Salt lowers the freezing temperature. (That is, it can remain a liquid at much lower temperatures.)
When sprinkled on ice, the salt lowers the freezing temperature of the water which effectively melts the ice when the salt dissolves into it. There is a limit to how low it can reduce the temperature, though. If the temperature drops below -9°C (15°F), it's too cold for the salt to dissolve into the ice.
When making ice cream, the salt lowers the temperature of the ice and water sufficiently enough to freeze the cream.