Answer:
5.97×10^24 kg
Explanation:
thats how it is expressed
Answer: The question is incomplete or some details are missing. Here is the complete question ; (a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.55 m/s2 for 4.05 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? m/s
(b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2
a ) With what speed (in m/s) does the car then strike the tree? m/s = 4.3125m/s
b) then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2 = -5.696m/s2
Explanation:
The detailed steps and calculation is as shown in the attached file.
Answer:
C
Explanation:
From the question we are told that a vector on the x and y plane face their negative axis
Generally in the x and y plane thr negative y axis is made to face down opposite the positive y axis
Whilst the negative x axis faces the left which is also alternate to the positive x axis
Generally A vector pointing towards the x and y negative axis fro the origin (0) will definitely be in the third quadrant
Answer:

Explanation:
Given data
Radius R=6.80 m
Velocity v=18 km/h =5 m/s
First we need to set up for Force in vertical (y) and horizontal (x)
∑Fy=0=TCosα-W
∑Fx=ma=Fc=TSinα
Solve Vertical force equation for T,substituting mg for W

Substitute expression for Fc and T into horizontal force and simplify it we get