Answer:
To be able to eat the food readily available in the environment
F=ma
8480=26.5m
m=8480/26.5
m=320
The mass of the cart is 320kg.
Question
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g
Part (b) what is the angle of banking of the highway? Give your answer in degrees
Answer:
a. Equation of Tangent
tan(θ) = v²/rg
b. Angle of the banking highway
θ = 0.087°
Explanation:
Given
Radius of the curve = r = 330m
Acceleration of gravity = g = 9.8m/s²
Velocity = v = 8km/h = 8 * 1000/3600
v = 2.22 m/s
a . Write an equation for the tangent of the highway's angle of banking
The Angle is calculated by
tan(θ) = v²/rg
θ = tan-1(v²/rg)
b.
Part (b) what is the angle of banking of the highway? Give your answer in degrees
θ = tan-1(v²/rg)
Substituting the values of v,g and r
θ = tan-1(2.22²/(330 * 9.8)
θ = tan-1(0.001523933209647)
θ = 0.087314873580116°
θ = 0.087°
The formula for the density of a substance expressed in mass and volume is rho = mass/volume or p = m/v. Rearranging the formula to isolate volume gives the formula v = m/p. To solve for the problem given, this formula must be used. This gives a solution of:
v = m/p = 250 g/ 968 g/cm^3 = 0.258 cm^3 of sodium
Answer:
Bank angle = 35.34o
Explanation:
Since the road is frictionless,
Tan (bank angle) = V^2/r*g
Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2
Tan ( bank angle) = 40^2/(230*9.81)
Tan (bank angle) = 0.7091
Bank angle = tan inverse (0.7091)
Bank angle = 35.34o
