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storchak [24]
3 years ago
7

A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c

ar travel if there is to be low tendency to skid even on very slippery pavement​
Physics
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

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8,961

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Which component of the earth’s atmosphere is decreased due to photosynthesis?
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A 40kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hill side?
bagirrra123 [75]

Answer:

<h2>39.2 m</h2>

Explanation:

The height of the hill side can be found by using the formula

h =  \frac{p}{m}  \\

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<h3>39.2 m</h3>

Hope this helps you

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According to Newton’s First Law of Motion, what will happen to a baseball thrown in outer space
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A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of
spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

                                       Q =  m×c×ΔT

                        Heat gain by aluminium + heat lost by copper  = 0    (1)

For Aluminium:

      Q = m\times0.897\frac{J}{g.k}\times(25-5)

      Q = m x 17.94 joule

For Copper:

Q= 206g\times0.385\frac{J}{g.k} \times(88-25)

       Q= 4996.53 Joule

from eq 1

     m x 17.94 = 4996.53

     mass of aluminium = \frac{4996.53}{17.94} g

    Mass of the aluminium chunk = 278.51 g

                         

3 0
3 years ago
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