You will have to use this formula:
Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs
Then:
-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2
Ps: It's value is negative because the she was in retrograde motion.
Answer: Her acceleration is -2 m/s^2.
Answer:
t=0.704s
Explanation:
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?
from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed
Vy^2=0+2*9.8*1.73sin40.5
Vy^2=22.021
Vy=4.69m/s
Vx^2=u^2+2*9.81*cos40.5
Vy^2=25.81
Vy=5.08m/s
V=(Vy^2+Vx^2)^0.5
V=47.71^0.5
V=6.9m/s
from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.
f=force applied
v=velocity final
u=initial velocity
m=mass of the toy, 0.046
f=ma
f=m(v-u)/t
v=u+at
6.9=0+9.8t
t=6.9/9.81
t=0.704s
Answer:
a) W = 643.5 J, b) W = -427.4 J
Explanation:
a) Work is defined by
W = F. x = F x cos θ
in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero
W = F x
let's calculate
W = 165 3.9
W = 643.5 J
b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes
sin θ = Wₓ / W
cos θ = Wy / W
Wₓ = W sinθ = mg sin θ
Wy = W cos θ
the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero
W = Wₓ x cos 180
W = - mg sin 34 x
let's calculate
W = -20 9.8 sin 34 3.9
W = -427.4 J
The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.
A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exerted on the system are equal in magnitude. The following statements predicts the motion of the skydiver at this time
<u>The skydiver is moving downward with constant speed.</u>
Explanation:
Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.
The forces acting on a falling leaf are : gravity and air resistance.
The net force and the acceleration on the falling skydiver is upward.
An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.
As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.
<u>A skydiver falling at a constant speed opens his parachute. When the skydiver is falling, the forces are unbalanced.</u>
Answer:
option D.
Explanation:
The correct answer is option D.
When an object is in equilibrium torque calculated at any point will be equal to zero.
An object is said to be in equilibrium net moment acting on the body should be equal to zero.
If the net moment on the object is not equal to zero then the object will rotate it will not be stable.
