Answer:
k = 1 700.7 N/m
v0 = 9.8 m/s^2
Explanation:
Hello!
We can answer this question using conservation of energy.
The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.
When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.
PS = (1/2) k x^2 <em>where x is the compresion or elongation of the spring</em>
PG = mgh
a)
Since energy must be conserved and we are neglecting any energy loss:
PS = PG
Solving for k
k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)
k = 1 700.7 N/m
b)
Since the potential energy of the spring transfors to kinetic energy of the ball we have that:
PS = KE
that is:
(1/2) k x^2 = (1/2) m v0^2
Solving for v0
v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)
v0 = 9.8 m/s^2
Answer:
a
b
c
Explanation:
From the question we are told that
The mass of the bag is 
The normal force experienced is 
The maximum acceleration of the bag is 
Generally this normal force experience by the bag is mathematically represented as
=> 
=> 
=> ![\theta = cos^{-1}[0.9183]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20cos%5E%7B-1%7D%5B0.9183%5D)
=>
Generally for the bag not to slip , it means that the frictional force is equal to the sliding force
Hence 
is mathematically represented as
While 
is mathematically represented as
So
=> 
=>
Generally from the workdone equation we have that
Here 
is the work done by friction which is mathematically represented as
Here s is the distance covered by the bag
is zero given that velocity at rest is zero
and
so
=> 
substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that
=>
Answer:
Δ P = 13.24 Pa
Explanation:
Given that
Density of oil ,ρ₁ = 9 x 10² kg/m³
We know that density for mercury ,ρ₂ = 13.6 x 10³ kg/m³
The change in the height of column ,Δh = 1.5 mm
The pressure given as
P = ρ g h
Change in the pressure
Δ P = ρ₁ g Δh
Now by putting the values
Δ P = 9 x 10² x 9.81 x 1.5 x 10⁻³ Pa
Δ P = 13.24 Pa
Therefor the change in the pressure will be 13.24 Pa.
Two methods of transfer of heat are involved in this process: conduction and convection.
In fact, the metal spoon is heated by conduction because the molecules of the boiling water collide with the molecules of the spoon, releasing heat to it; and also by convection, because in the pot of boiling water masses of hot water goes upward and they give their heat to the spoon, then these masses become cooler and they go down, replaced by other masses of hot water.
