The radius of the curved road at the given condition is 54.1 m.
The given parameters:
- <em>mass of the car, m = 1000 kg</em>
- <em>speed of the car, v = 50 km/h = 13.89 m/s</em>
- <em>banking angle, θ = 20⁰</em>
The normal force on the car due to banking curve is calculated as follows;
The horizontal force on the car due to the banking curve is calculated as follows;
<em>Divide </em><em>the second equation by the first;</em>
Thus, the radius of the curved road at the given condition is 54.1 m.
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The vehicle's centripetal acceleration is equal to 22.5m/s²
Radius, r = 10 meter
Speed, V = 15 m/s
To ascertain the car's centripetal acceleration
A(c) = V²/R
We obtain the following when we enter the formula's parameters:
A(c) = 152/10
A(c) = 225/10
A(c) = 22.5m/s²
<h3>What is Centripetal acceleration ?</h3>
When an item moves in a circular route, one of its motion characteristics is centripetal acceleration. Any motion in a circle with an acceleration vector pointing in the direction of the circle's centre is referred to as centripetal acceleration.
- Centripetal forces cause accelerations at the centripetal axis. With the exception of the Earth's rotation around the Sun, any satellite's circular motion around a celestial body is brought on by the centripetal force produced by their mutual gravitational pull.
Hence, Centripetal acceleration is
22.5 m/s²
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The watch hand covers an angular displacement of 2π radians in 60 seconds.
ω = 2π/60
ω = 0.1 rad/s
v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
The area of the Earth (Ae) that is irradiated by is given by:
Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2
On the Earth, insolation (We) = Psun/Ae
Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W
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