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.
Learn more about banking angle here: brainly.com/question/8169892
Answer:
H = 0.673
Explanation:
given,
side of cubical crate = 0.74
weight of the crate = 600 N
magnitude of force = 330 N
the Horizontal distance of its Center of mass
= 0.74/2
= 0.37
Let the required Height be H
By Balancing the Torques, we get
H x 330 N = 0.37 x 600
330 H = 222
H = 0.673
hence, the height above the floor where force is acting is equal to 0.673 m
Answer:
here given is a weight
then force becomes mg
that is F=Mg
=4*9.8
then by using the formula
F=Ma
a=F/M
=4*9.8/9.8
=4
Explanation:
From p1v1/t1 = p2v2/t2
pressure unchanged ... cancelled out
v1=605 , t1=27C = 300K,
t2=-3C = 270K
***remember temperature must be in Kelvin
we got
605/300 = v2/270
v2 = 545
