Answer:
The normal force experienced by the car is approximately 8223.2 N
Explanation:
The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction
The mass of the vehicle that moves around the curve, m = 900 kg
The incline of the curve, θ = 20°
The speed with which the vehicle moves around the curve, v = 12.5 m/s
The radius of the curve, R = 50 meters
We have;
![N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}](https://tex.z-dn.net/?f=N%20%5Ccdot%20sin%28%5Ctheta%29%20%3D%20%5Cdfrac%7Bm%20%5Ccdot%20v%5E2%7D%7BR%7D)
Where;
θ = The angle of inclination of the road = 20°
N = The normal force experienced by the car
m = The mass of the car = 900 kg
v = The velocity with which the car is moving = 12.5 m/s
R = The radius of the curve around which the vehicle moves = 50 m
![\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})} = 8223.1998754586828969046217875927](https://tex.z-dn.net/?f=%5Ctherefore%20N%20%3D%20%5Cdfrac%7Bm%20%5Ccdot%20v%5E2%7D%7BR%20%5Ccdot%20sin%28%5Ctheta%29%7D%20%3D%20%5Cdfrac%7B900%20%5Ctimes%20%2812.5%29%5E2%7D%7B50%20%5Ctimes%20sin%2820%5E%7B%5Ccirc%7D%29%7D%20%20%3D%208223.1998754586828969046217875927)
The normal force experienced by the car = N ≈ 8223.2 N.
Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m
But cracked cracks nuts while scripts cut
Explanation:
It is given that,
The total mass of the wagon, rider and the rock is 99 kg
The mass of the rock is 0.261 kg.
The initial speed of the wagon, u = 0.519 m/s
Speed of the rock, u' = 15.9 m/s
(a) Let v is the speed of the wagon after the rock is thrown directly in forward direction.
The momentum will remain constant. Using the conservation of momentum as :
![99\times 0.261=0.261\times 15.9+(99-0.261)v](https://tex.z-dn.net/?f=99%5Ctimes%200.261%3D0.261%5Ctimes%2015.9%2B%2899-0.261%29v)
v = 0.22 m/s
(b) Let v' is the speed of the wagon after the rock is thrown directly in opposite direction, u' = -15.9 m/s
![99\times 0.261=0.261\times (-15.9)+(99-0.261)v](https://tex.z-dn.net/?f=99%5Ctimes%200.261%3D0.261%5Ctimes%20%28-15.9%29%2B%2899-0.261%29v)
v' = 0.304 m/s
Hence, this is the required solution.
Displacement vectors of 6 km South 2 km North, 7 km South, and 5 km North combine to a total displacement of 6 km South.
Answer: Option A
<u>Explanation:</u>
The displacement vector represents the location change: the distance of separation from the start point to the end point is the displacement vector’ magnitude, and travelled direction denotes the displacement vector’ direction.
In figure, the illustration shows a new vector for the entire journey from beginning to end. In other words, C = A + B. The C vector is called the sum, or resultant vectors. When applying this concept to the given question, we can find the total displacement vector value. It is as follows:
Given:
6 km South 2 km North, 7 km South, and 5 km North
Total displacement, s = 6 km South - 2 km North + 7 km South - 5 km North
North and south lie in an opposite direction. Therefore, when combining all distances negative sign mentioned to denote the direction.
Total displacement, s = 13 km South - 7 km North
Total displacement, s = 6 km South