Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
<span>Wt. of skier. </span>
<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>
<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>
<span>Ek + Ep = Ekmax - Fk*d </span>
<span>Ek = Ekmax-Ep-Fk*d </span>
<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>
To solve this problem we will apply the relationship between Newton's second law and Hooke's law, with which we will define the balance of the system. From the only unknown in that equation that will be the constant of the spring, we will proceed to find the period of vibration of the car.
We know from Hooke's law that the force in a spring is defined as
Here k is the spring constant and x the displacement
While by Newton's second law we have that the Weight can be defined as
Here m is the mass and g the gravity acceleration.
The total weight would be
Each spring takes a quarter of the weight, then
Since the system is in equilibrium the force produced by the weight in each spring must be equivalent to the force of the spring, that is to say
The period of a spring-mass system is given as
The total mass is equivalent as the sum of all the weights, then replacing we have that the Period is
Therefore the period of vibration of the car as it comes to rest after the four get in is 0.9635s
Answer:
v₀ = 14.6 m / s
Explanation:
This is a projectile launch exercise, in this case we must stop the wall, we can assume that at this point the ball is at its maximum height
as the height is maximum its vertical speed is zero
v_{y} = - g t
y = v_{oy} t - ½ g t²
the horizontal distance is
x = v₀ₓ t
we use trigonometry to find the components of the velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v_{o} sin θ
we substitute the values
0 = v₀ sin θ - g t
10 = v₀ sin θ t - ½ g t²
6 = v₀ cos θ t
we have three equations and three unknowns by which the system can be solved
let's use the first and second equations,
10 = (gt) t - ½ g t²
10 = g t² - ½ g t²
10 / 9.8 = t² (1- ½)
1.02 = ½ t²
t = √ 2.04
t = 1,429 s
this is the time to hit the wall
now let's use the first and third equations, we substitute the data
v₀ sin θ = g t
v₀ cos θ = x / t
we divide and simplify
tan θ = g t² / x
θ = tan⁻¹ g t² / x
θ = tan⁻¹ (9.8 1.429²/6)
θ = 73.3°
we substitute in the third equation
x = v₀ cos 73.3 t
v₀ = x / t cos 73.3
v₀ = 6 / (1.429 cos 73.3)
v₀ = 14.6 m / s
Answer:
The time is
Explanation:
From the question we are told that
The angle is
The height is
Generally from SOHCAHTOA the length of the ramp can be evaluated as
=>
=>
Generally fro kinematic equation this length is mathematically represented as
Here u = 0 m/s since the box started from rest
So
Answer:
For ball P=- 2 x m x V.
For P= - m x V
Explanation:
As we know that
if external force is zero then linear momentum of system will be conserve.
Rubber ball:
Given that speed of ball is V before the collision with ball.
The speed of the ball is remain after the collision with ball but the direction will reverse.
Lets take first ball is moving in positive x direction and after collision with ball it moves in negative x direction.
Initial momentum
Finall momentum
So the change in the momentum
P=- 2 x m x V.
For clay:
Initial velocity of clay is V in positive x direction and final velocity is zero because it stick with the wall
So the change in the momentum
P= - m x V.