Earths atmosphere, Temperature, Plant life and Oxygen I guess
Answer:
w = 0.943 rad / s
Explanation:
For this problem we can use the law of conservation of angular momentum
Starting point. With the mouse in the center
L₀ = I w₀
Where The moment of inertia (I) of a rod that rotates at one end is
I = 1/3 M L²
Final point. When the mouse is at the end of the rod
= I w + m L² w
As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved
L₀ = L_{f}
I w₀ = (I + m L²) w
w = I / I + m L²) w₀
We substitute the moment of inertia
w = 1/3 M L² / (1/3 M + m) L² w₀
w = 1 / 3M / (M / 3 + m) w₀
We substitute the values
w = 1/3 / (1/3 + 0.02) w₀
w = 0.943 w₀
To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s
w = 0.943 rad / s
Answer:

Explanation:
The inital electrostatic force between the two spheres is given by:

where
is the initial force
k is the Coulomb's constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have

Therefore, the new electrostatic force will be

So the force has increased by a factor 4. By using
, we find

Answer:
Impulse = 88 kg m/s
Mass = 8.8 kg
Explanation:
<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>
Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:
Eqn. (1)
where
: is the Net Force in Newtons (
)
: is the mass (
)
: is the acceleration (
)
We also know that the acceleration is denoted by the velocity (
) of an object as a function of time (
) with
Eqn. (2)
Now substituting Eqn. (2) into Eqn. (1) we have
Eqn. (3)
However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e.
is in fact the Impulse
of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈
and
(thus just before the cut-off time of the force acting).
Thus to find the Impulse we have:

So the impulse of the cart is 
Then, we know that the cart is moving at
. Plugging in the values in Eqn. (3) we have:

So the mass of the cart is
.