<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
Answer:
round
Explanation:
science explained it alot
You can use Vf^2-Vi^2 = 2ax
Vf^2 - 0 = 2(9.81)(25)
Or you can use energy
mgh = 1/2mv^2
2gh =v^2
Same thing
Answer:
Explanation:
The vehicle is experiencing a large force created by the concrete wall.
Equation
vf^2 = vi^2 + 2*a * d
Givens
vf = 0 The car eventually does stop.
vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]
vi = 20 meters / second
a = ?
m = 850 kg
Solution
vf^2 = vi^2 + 2a*d
0 = 20 m/s + 2* 2 *a
-20 m/s = 4a
-20/4 = a
a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.
Force = m * a
F = - 850 * (-5)
F = - 4250 N
The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.
Answer:
F = 789 Newton
Explanation:
Given that,
Speed of the car, v = 10 m/s
Radius of circular path, r = 30 m
Mass of the passenger, m = 60 kg
To find :
The normal force exerted by the seat of the car when the it is at the bottom of the depression.
Solution,
Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.
N = 788.6 Newton
N = 789 Newton
So, the normal force exerted by the seat of the car is 789 Newton.
