Answer:
Explanation:
The speed in meters per second is ...
(90 km/h)(1000 m/km)(1 h/(3600 s)) = 25 m/s
The braking acceleration can be found from ...
a = v²/(2d) = (25 m/s)²/(2×50 m) = 6.25 m/s²
Then the braking force is ...
F = Ma = (1800 kg)(6.25 m/s²) = 11,250 N
The torque on the center of gravity is ...
T = (11,250 N)(0.90 m) = 10,125 N·m
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If we let x and y represent the normal forces on the front and rear wheels, respectively, then we have ...
x + y = (10 m/s²)(1800 kg) = 18000 . . . . . newtons
1.7x -1.3y = 10,125 . . . . . . . . . . . . . . . . . . . newton-meters
The latter equation balances the torque due to the wheel normal forces with the torque due to braking forces.
Multiplying the first equation by 1.3 and adding that to the second, we have ...
3.0x = (1.3)(18,000) + 10,125
x = 33,525/3 = 11,175 . . . . . . . . . . . newtons normal force on front tyres
y = 18000 -11175 = 6,825 . . . . . . . .newtons normal force on back tyres
The least coefficient of friction is the ratio of horizontal to vertical acceleration, 6.25/10 = 0.625.
Answer:
Explanation:
The strength of the gravitational field at the surface of a planet is given by
(1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
For the Earth:
For the unknown planet,
Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:
And substituting g = 9.8 m/s^2,