Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²
= (1/2) (9.8 m/s²) (95 sec)²
= (4.9 m/s²) (9,025 sec²)
= 44,222.5 meters .
The depth of the mine shaft is five times the height of Mt. Everest !
Directly proportional to pressure
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
Answer:
Explanation:
This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the car in this problem,
u = 27.8 m/s
v = 0
s = 17 m
Solving for a, we find the acceleration:
Answer
given,
mass of the drop, m = 0.0014 g
speed of the drop, u = 8.1 m/s
a) Change in momentum is equal to impulse
final velocity of the drop, v = 0 m/s
J = m ( v - u )
J = 0.0014 x 10⁻³ x ( 0 - 8.1 )
J = -1.134 x 10⁻⁵ kg.m/s
impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s
b) time, t = 0.37 m s
impact of force = ?
we know
J = F x t
1.134 x 10⁻⁵ = F x 0.37 x 10⁻³
F = 0.031 N
the magnitude of the force of the impact is equal to F = 0.031 N
