Answer:
The answer to the question is
The correct weight of the former student = 116 lbf
Explanation:
To solve the question, we list out the known variables as follows
Capacity of the scale = 120 lbs
Capacity of the spring dynamometer = 20 lbs
Reading on the scale = 100lb
Reading on the spring dynamometer= 16 lb
at equilibrium, sum of forces = 0
Therefore weight of former student = reaction forces of the scale + the reaction force of the spring dynamometer = 100lb + 16 lb = 116 lb
The correct weight = 116lb weight or 52.62 kg
Converting to weight which is a force = mass * acceleration = 52.62 kg * 9.81 m/s² =
516.2 N = 116 lbf
Answer:
equal
Explanation:
If all three corresponding sides are equal and all three corresponding angles are identical in measure, two triangles are said to be congruent. These triangles can be slides, rotated, flipped and turned to be looked identical. If repositioned, they coincide with each other. The symbol of congruence is' ≅'.
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)
Answer:
Explanation:
Given that:
- moment of inertia of tucked body,
- rotational speed of the body,
- i.e.
- moment of inertia of the straightened body,
<u>Now using the law of conservation of angular momentum:</u>
angular momentum of tucked body=angular momentum of straight body
Your potential energy at the top of the hill was (mass) x (gravity) x (height) .
Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .
If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.
(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)
Divide each side by 'mass' :
(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.
Multiply each side by 2 :
(speed)² = 2 x (gravity) x (height)
Take the square root of each side:
<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>
We could go one step further, since we know the acceleration of gravity on Earth:
Speed at the bottom = 4.43 x square root of (height of the hill)
This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom. The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.
