The pressure at the bottom of a glass filled with water (r 5 1 000 kg/m3 ) is P. The water is poured out and the glass is filled
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Answer:
top speed = 17.25
Total height = 281.19 m
Explanation:
given data
mass = 75 kg
thrust = 160 N
coefficient of kinetic friction = 0.1
solution
we get here frictional force acting that is
frictional force = .............1
frictional force = 0.1 × 75 × 9.8
frictional force = 73.5 N
and
Net force acting will be F = 160 - 73.5 N
F = 86.5 N
so
Acceleration in the First 15 second will be
F = ma .........2
86.5 = 75 × a
a = 1.15 m/s²
and
now After 15 second the velocity will be as
v = u + at ..........3
here u is 0
so v will be
V = 1.15 × 15
v = 17.25
and
now we get travels distance S in 15 s
s = u × t + 0.5 × a × t²
here u is 0
so distance s will be
s = 0.5 × a × t²
s = 0.5 × 1.15 × 15²
s = 129.37 m
and
now acceleration acting is
F =
m a =
a =
a = - 0.98
here it is negative it mean downward nature of acceleration
and
now we get distance s by this formula
V² - u² = 2 a s
here v velocity is 0 and
u initial velocity is 17.25 m/s
put here value
0 - 17.25² = 2 × (-0.98) × s
solve it we get
s = 151.82 m
so
Total height is
Total height = 129.37 m + 151.82 m
Total height = 281.19 m
The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
Which gives;
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
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