A = 59.35cm
B = 196.56g
C = 74.65g
<u>Explanation:</u>
We know,
and L = x+y
1.
Total length, L = 100cm
Weight of Beam, W = 71.8g
Center of mass, x = 49.2cm
Added weight, F = 240g
Position weight placed from fulcrum, y = ?
Therefore, position weight placed from fulcrum is 59.35cm
2.
Total length, L = 100cm
Center of mass, x = 47.8 cm
Added weight, F = 180g
Position weight placed from fulcrum, y = 12.4cm
Weight of Beam, W = ?
Therefore, weight of the beam is 196.56g
3.
Total length, L = 100cm
Center of mass, x = 50.8 cm
Position weight placed from fulcrum, y = 9.8cm
Weight of Beam, W = 72.3g
Added weight, F = ?
Therefore, Added weight F is 74.65g
A = 59.35cm
B = 196.56g
C = 74.65g
Spring is stretched by force f to distance "x"
now here by force balance we can say
now here we will we say that energy stored in the spring will convert into kinetic energy
now solving above equation we will have
PART 2)
now for half of the extension again we can use energy conservation
now the speed is given as
Answer:
The age of the universe would be 9.9 billion years
Explanation:
We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.
Thus, the time it has taken for the galaxies to reach their current separations is:
and from Hubble's Law:
Therefore:
With the given value for the Hubble's constant we have:
and thus,
Answer:
m = 0.01 kg
Explanation:
Given that,
Momentum of the marble, p = 0.15 kg-m/s
Speed of the marble, v = 15 m/s
We need to find its mass. We know that,
Momentum, p = mv
Where
m is the mass
So, the mass of the marble is equal to 0.01 kg.
Answer:
Explanation:
Moment of inertia of the metal rod pivoted in the middle
= M l² / 12
If the spring is compressed by small distance x twisting the rod by angle θ
restoring force by spring
= k x
moment of torque about axis
= k x l /2
= k θ( l /2 )² ( x / .5 l = θ )
=
moment of torque = moment of inertia of rod x angular acceleration
k θ( l /2 )² = M l² / 12 d²θ/dt²
d²θ/dt² = 3 k/M θ
acceleration = ω² θ
ω² = 3 k/M
ω = √ 3 k / M
