Answer:
a) 25.5°(south of east)
b) 119 s
c) 238 m
Explanation:
solution:
we have river speed =2 m/s
velocity of motorboat relative to water is =4.2 m/s
so speed will be:
a) =
+
solving graphically
=4.7 m/s
Ф=
=25.5°(south of east)
b) time to cross the river: t==
=119 s
c) d==(2)(119)=238 m
note :
pic is attached
Answer:
Explanation:
Given
Mass of the elevator is
Time period of ascension
cruising speed
Distance moved by elevator during this time
Suppose Elevator starts from rest
Distance moved
Gain in Potential Energy is
Average power during this period is
Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load = 10000 kg
mass of flat bed = 20000 kg
initial speed of truck = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs -------------let this be equation 1
where = normal force = mg
so
Fs,max = μs mg
ma = μs mg
divide through by mass
a = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a = 0.5 × 9.8 m/s²
a = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
² =
² + 2aΔx
where is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load