Answer:
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Explanation:
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Answer:
<h2>The coefficient of static friction will be
0.7</h2>
Explanation:
Given data
the radius of curve= 90m
speed v= 90 km/h to m/s = (90*100)/60*60= 25 m/s
we know that the expression for the centripetal force acting on the car
-------1
we also know that the expression for the frictional force between road and tire.
Ff= μmg--------2
Equating equation 1 and 2 we have
μmg= mv^2/r
μ= v^2/gr
substituting the values of speed and radius we have (assuming g= 9.81m/s^2)
μ= 25^2/9.81*90
μ= 625/882.9
μ= 0.7
Answer:
v_f = 6.92 x 10^(4) m/s
Explanation:
From conservation of energy,
E = (1/2)mv² - GmM/r
Where M is mass of sun
Thus,
E_i = E_f will give;
(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)
m will cancel out to give ;
(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)
Let's make v_f the subject;
v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]
G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²
Mass of sun is 1.9891 x 10^(30) kg
v_i = 2.1×10⁴ m/s
r_i = 2.5 × 10^(11) m
r_f = 4.9 × 10^(10) m
Plugging in all these values, we have;
v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12
v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]
v_f = √[(441000000) + (435.38 x 10^(7))
v_f = 6.92 x 10^(4) m/s
It should be Waterlogged streets
