Formulas you need for this problem:
F= mass•acceleration
KE= mass•velocity^/2
Acceleration= final velocity-intial velocity/time
time= distance/speed
t= 97.5\10= 9.75 seconds
Acc= 10-20/9.75= -1.03 m/s/s
These are the answers :)
F=650•-1.03= -669.5N
KE= 650•100/2= 32,500J or 32.5KJ
Answer:
W = y (b-a) / ab
Explanation:
Work is defined by the expression
W = ∫ F. dr
In this case the force is in the same direction of displacement, so the scalar product is reduced to the ordinary product
W = ∫ F dr
The expression of the strength left is
F = -y / x²
let's replace and integrate
W = ∫ (-y / x²) dx
W = -y (-1 / x)
We evaluate between the lower limit x = b + a to the upper limit x = 0 + a
W = -y (-1 / b + 1 / a)
W = y (b-a) / ab
where (b-a) is the distance traveled
Answer:
Gravitational force between masses there is Gravitational force .
Explanation:
Using equation of motion to determine the acceleration of the car,
vf^2 = vi^2 + 2 * a * S,
vf = 0
0 = vi^2 + 2 * a * S
Converting mph to m/s,
3 mph * 5280 ft/mi * 12 in/ft * 2.54 cm/in * 1 m/100 cm * 1 h/3600 s
= 3 * 0.445
v = 1.335 m/s
Converting in to m,
2 in * 2.54 cm/in * 1 m/100 cm = 0.0254 m
= 2 * 0.0254
S = 0.0508 m
0 = 1.335^2 + 2 * a * 0.0508
a = -1.335^2 ÷ 0.1013
= -17.54 m/s^2
Mass of car (assumed) = 2000 kg
Force = ma
= 2000 × 17.54
= 35.08 kN.
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