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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?
Physics
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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maxonik [38]

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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3 years ago
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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
Dimas [21]

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

4 0
3 years ago
Biologists use ball-and-stick models to study complex molecules by representing atoms with balls and bonds with sticks. DNA is a
AleksAgata [21]
One of the major limitations of using the ball and stick model for DNA, is that within a single double stranded segment of DNA, one would have to use many many balls to represent atoms that are present in the sugar phosphate backbone, along with all of the main atoms that compose the nitrogenous bases of DNA, we also cannot construct or show the helical form of DNA, by using balls and sticks as well.
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3 years ago
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}

6 0
3 years ago
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allsm [11]

Answer:

D) Grounding

Explanation:

The potential difference between cloud and ground leads to ionization of the atmosphere and resulting conduction through the air often to ground (although it can be between clouds at different potentials. I would say grounding, like the spark when you touch a hot battery terminal to ground on a car.

5 0
3 years ago
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