Answer:
The height of the cliff is, h = 78.4 m
Explanation:
Given,
The horizontal velocity of the projectile, Vx = 20 m/s
The range of the projectile, s = 80 m
The projectile projected from a height is given by the formula
<em> S = Vx [Vy + √(Vy² + 2gh)] / g
</em>
Therefore,
h = S²g/2Vx²
Substituting the values
h = 80² x 9.8/ (2 x 20²)
= 78.4 m
Hence, the height of the cliff is, h = 78.4 m
Answer:
Let's investigate the case where the cable breaks.
Conservation of angular momentum can be used to find the speed.
The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>
where r is the length of the cable.
<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is
The relation between linear velocity and the angular velocity is
So,
As can be seen, the maximum velocity for the projectile without breaking the cable is 
, where r is the length of the cable.
Answer:
x(t) = d*cos ( wt )
w = √(k/m)
Explanation:
Given:-
- The mass of block = m
- The spring constant = k
- The initial displacement = xi = d
Find:-
- The expression for displacement (x) as function of time (t).
Solution:-
- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.
- We apply the Newton's equation of motion in horizontal direction.
F = ma
-kx = ma
-kx = mx''
mx'' + kx = 0
- Solve the Auxiliary equation for the ODE above:
ms^2 + k = 0
s^2 + (k/m) = 0
s = +/- √(k/m) i = +/- w i
- The complementary solution for complex roots is:
x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]
- The given initial conditions are:
x(0) = d
d = [ A*cos ( 0 ) + B*sin ( 0 ) ]
d = A
x'(0) = 0
x'(t) = -Aw*sin (wt) + Bw*cos(wt)
0 = -Aw*sin (0) + Bw*cos(0)
B = 0
- The required displacement-time relationship for SHM:
x(t) = d*cos ( wt )
w = √(k/m)
There are 92 natural ones and about 20 more made in particle-Physics labs.
If we only stick to the 92 natural ones, then there are
(92!) / (88! times 4!) = 2,794,155 possible answers to your question.
Here's one: Osmium
Neodymium
Iridium
Gold .
