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vagabundo [1.1K]

3 years ago

9

If a curve with a radius of 65 m is properly banked for a car traveling 75 km/h, what must be the coefficient of static friction

for a car not to skid when traveling at 104 km/h?

1 answer:

STALIN [3.7K]3 years ago

8 0

As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. Equation to follow are F = mv²/r "Force of friction" Ff= μsN, Normal N=mg Ff=μsmg. Therfore
mv²/r = μsmg cancel mass "m" The formula to follow will be μs = v²/rg

μs = v²/rg μs = (28.89 m/s²)²/ (65 m)(9.8 m/s²) = 1.31

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Answer:

answer is 0.1428

Explanation:

Data:- vf=5.0 , vi=0.0 , t=35 , a=? so appling first eq of motion vf=vi+at we have to find a=vf-vi/t , a=5.0-0.0/35 , a=5/35 ,a=0.1428m/sec²

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Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120

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Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) $\left | a\right |=5\ km$

the magnitude of the second vector(say) $\left | b\right |=7\ km$

the angle between them is $120^{\circ}$

The resultant vector magnitude is given by

$\left | \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}$

$\left | \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left | \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left | \vec{R}\right |=6.24\ km$

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A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

a)

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

b)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

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allsm [11]

Answer:

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