Answer:Explanation:
Image result for what does a worm and wheel mechanism do to torque and speed
Like other gear arrangements, a worm drive can reduce rotational speed or transmit higher torque. ... Each full 360 degree turn of a single start worm advances the gear by one tooth. For a multi start worm the gear reduction equals the number of teeth on the gear divided by the number of starts on the worm.
The answer, using an indicator to measure the hydrogen ion concentration of a solution, is correct
The maximum force that the tires can exert on the road before slipping is 16200 N.
From the information in the question;
The coefficient of static friction = 0.9
The mass of the car = 1800 kg
Using the formula;
μ = F/R
μ = coefficient of static friction
F = force on the tires
R = the reaction force
But recall that the reaction is equal in magnitude to the weight of the car.
W=R
Hence; R = 1800 kg × 10 ms-2 = 18000 N
Making F the subject of the formula;
F = μR
Substituting values;
F = 18000 N × 0.9
F = 16200 N
Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.
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A :-) a = v^2 by r
Given - radius = 25 m
velocity = 10 m/s
Solution -
a = v^2 by r
a = ( 10 )^2 by 25
a = 100 by 25
( cut 25 and 100 because 25 x 4 = 100 )
a = 4 m/s^2
.:. The centripetal acceleration of the car
= 4 m/s^2.
Answer:
83.055 m
Explanation:
According to the given scenario, the calculation of skid distance is shown below:-

Where
u = 11.3
v = 0
t = 14.7
Now placing these values to the above formula,
So,

= 83.055 m
Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question