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katovenus [111]
3 years ago
5

A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.8

2, what is the greatest speed the car can have in the corner without skidding?
Physics
1 answer:
Vaselesa [24]3 years ago
5 0

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

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Here mass number on the product side is given to us

so sum of mass number is given as

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now on the reactant side also the number must be same

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8 0
2 years ago
A 6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is
ivanzaharov [21]
<span>anwser will be 

F = ma

where 

F = force exerted on the bullet 
m = mass of the bullet = 5 gm (given) = 0.005 kg. 
a = acceleration of the bullet 

Substituting appropriately, 

F = 0.005a --- call this Equation 1 

Next working equation is 

Vf^2 - Vo^2 = 2as 

where 

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given) 
Vo = initial velocity of bullet = 0 
a = acceleration of bullet 
s = length of the rifle's barrel 

Substituting appropriately, 

326^2 - 0 = 2(a)(0.83) 

a = 64,022 m/sec^2 

the anwser will be
Substituting this into Equation 1, 

F = 0.005(64,022) 

F =320.11 Newtons 

Hope this helps. </span><span>
</span>
8 0
3 years ago
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