Answer:
A. 90 degrees clockwise rotation
Step-by-step explanation:
Of we have a coordinate axis (x,y), if this axis is rotated 90° clock wide, the resulting coordinate of the pre-image will be the coordinate (y -x). Note that the coordinates was swapped and then the new y coordinate negated.
Given the coordinate K(24, -15). If we rotate this clockwisely, first we swap the coordinate axis to have (-15, 24)
Them we will negate the new y coordinate axis to have;
K'(-15, -24)
Therefore the correct answer is 90° clockwise rotation.
Answer:
a) weight of the car = 2816,1 lbs
b) 2773 lbs
Step-by-step explanation:
The equilibrium force is 490 lbs. That force keep the car at rest, then
∑ Fy = 0 and ∑Fx = 0
Forces acting on the car:
The external force 490 lbs
weight of the car uknown
Normal force
sin∠10° = 0,174
cos∠10° = 0.985
∑Fx = 0 mg*sin10°- 490 = 0 ∑Fy = 0 mg*cos10° - N = 0
mg*0,174= 490
mg = 490 / 0,174
mg = 2816,1 lbs
weight of the car = 2816,1 lbs
The Normal force
mg*cos10° - N = 0 2816,1 * 0,985 = N
N = 2773 lbs
Then equal force in magnitude and in opposite direction will car exets on the driveway
Answer:
3:5
Step-by-step explanation:
5a + 3b = 6b
5a=6b-3b
5a=3b
a/b= 3/5
a:b=3:5