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ASHA 777 [7]
1 year ago
11

At what angle should the roadway on a curve with a 45 m radius be banked to allow cars to negotiate the curve at 12 m/s even if

the roadway is icy (and the frictional force is zero)?.
Physics
1 answer:
nordsb [41]1 year ago
8 0

At an angle of  18.26 °  the roadway on a curve with a 45 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is icy

angle of banking of the car going round the curve = tan (theta) = v^{2}  / rg

v = speed

r = radius

g = acceleration due to gravity

substituting the values

tan (theta)  = 12^{2} / 45 * 9.8  = 0.33  

theta = tan^{-1} (0.33)

         = 18.26 °

At an angle of  18.26 °  the roadway on a curve with a 45 m radius be banked to allow cars to negotiate the curve at 12 m/s even if the roadway is icy .

To learn more about acceleration due to gravity here :

brainly.com/question/13860566

#SPJ4

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A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
julsineya [31]

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15

Explanation:

mass, M = 5Kg

horizontal force, F_h = 40N

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frictional force, F_f =?

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4 0
3 years ago
A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the
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Answer:

1176.01 °C

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V = IR................. Equation 1

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make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

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2 years ago
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A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v
ch4aika [34]

Answer

given,

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v_max=1.00 m/s

we know,

v_ max=A ω  

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-0.50= -1.00 sin (ωt)

sin (ωt)  =  0.5

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t = 0.125 s

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= 0.50 sec

7 0
2 years ago
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