A compressed spring is compressed between a 5kg and 3 kg cart, respectively. When the spring is released, what will be the same
1 answer:
The momentum of the two carts will be the same
Explanation:
First of all, we notice that the system consisting of the two carts + the spring is an isolated system: this means that there are no external forces acting on it.
As a result, the total momentum of the system must be conserved.
Since the spring is at rest, the total momentum of the system after the two carts are released is:
where
is the momentum of the 1st cart
is the momentum of the 2nd cart
The total momentum of the system at the beginning, when the spring is compressed, is zero:
Since the total momentum must be conserved, we have:
Which means that the two carts will have equal and opposite momenta.
Learn more about momentum:
#LearnwithBrainly
You might be interested in
Heat required to raise the temperature of a given system is
here we know that
m = mass
s = specific heat capacity
= change in temperature
now as we know that
mass of wood = 5 kg
mass of aluminium pan = 2 kg
change in temperature = 45 - 20 = 25 degree C
specific heat capacity of wood = 1700 J/kg C
specific heat capacity of aluminium = 900 J/kg C
now here we will find the total heat to raise the temperature of both
So heat required to raise the temperature of the system is 257500 J
<h2>Δt = 71.67 °C</h2>
The temperature change of water is equal to 71.67 °C
<h3>Explanation:</h3>
Given:
Amount of transferred energy = 30,000 K J
Mass of water = 100 ml
Initial temperature = 20°C
To find the change in temperature of water.
Formula for Heat capacity is given by
Q = m×c×Δt ........................................(1)
where:
Q = Heat capacity of the substance (in J)
m=mass of the substance being heated in grams(g)
c = the specific heat of the substance in J/(g.°C)
Δt = Change in temperature (in °C)
Δt = (Final temperature - Initial temperature) = T(f) - T(i)
Q = 30,000 J
Mass of water = m = 100 ml
1 ml = 1 g ................................................(2)
Therefore m = 100 ml = 100 g
Specific heat of water is c = 4.186 J /g.
Δt = ?
Substituting these in equation (1), we get
Q = m×c×Δt
Rearranging the terms for Δt,
Δt =
Δt =
Δt = 71.67 °C