An equilibruium is not changed by a changed in pressure. the answer is false
Answer:
The kinetic energy of the particle will be 12U₀
Explanation:
Given that,
A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.
Constant force = 12F
According to question,
The kinetic energy is
....(I)
Constant force = 12F
A resistive force field is now set up ,
Resistive force is given by,

When the particle moves from point B to point A then,
We need to calculate the kinetic energy
Using formula for kinetic energy

Put the value of 

Now, from equation (I)

Hence, The kinetic energy of the particle will be 12U₀.
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.
The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.
v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s
Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.
The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.
Combining this together we get:
(1) vx=40m/s and vy=10m/s
Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w =
=
= 18.38 rad/sec
Period of oscillation = 
= 0.34 sec
hola ya sabes la respuesta si si ,me la dices porfa