Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
Answer:
Shiloh dynasty, jucie wrld or xxx or twenty one plot
Explanatio
The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
<h3>Average velocity of the car</h3>
The average velocity of the car is calculated as follows;
x(t) = a + bt + ct2
v = dx/dt
v(t) = b + 2ct
v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s
v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s
<h3>Average velocity</h3>
V = ¹/₂[v(0) + v(10)]
V = ¹/₂ (-10.1 + 11.9 )
V = 0.9 m/s
Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
Learn more about velocity here: brainly.com/question/4931057
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Answer:
<h3>2,321.62Joules</h3>
Explanation:
The formula for calculating workdone is expressed as;
Workdone = Force * Distance
Get the force
F = nR
n is the coefficient of friction = 0.5
R is the reaction = mg
R = 46 ( 9.8)
R = 450.8N
F = 0.5 * 450.8
F = 225.4N
Distance = 10.3m
Get the workdone
Workdone = 225.4 * 10.3
Workdone = 2,321.62Joules
<em>Hence the amount of work done is 2,321.62Joules</em>
