Answer:
The impulse on the object is 60Ns.
Explanation:
Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.
F = m a
F = m(
)
⇒ Ft = m(
-
)
where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object,
is its initialvelocity and
is the final velocity of the object.
Therefore,
impulse = Ft = m(
-
)
From the question, m = 3kg,
= 0m/s and
= 20m/s.
So that,
Impulse = 3 (20 - 0)
= 3(20)
= 60Ns
The impulse on the object is 60Ns.
Answer:
Time always is on X axis.
three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
<u>Answer:</u> The final temperature of the solution is 
<u>Explanation:</u>
The amount of heat released by coffee will be absorbed by aluminium spoon.
Thus, 
To calculate the amount of heat released or absorbed, we use the equation:

Also,
..........(1)
where,
q = heat absorbed or released
= mass of aluminium = 39 g
= mass of coffee = 166 g
= final temperature = ?
= temperature of aluminium = 
= temperature of coffee = 
= specific heat of aluminium = 
= specific heat of coffee= 
Putting all the values in equation 1, we get:
![39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]](https://tex.z-dn.net/?f=39%5Ctimes%200.904%5Ctimes%20%28T_%7Bfinal%7D-24%29%3D-%5B166%5Ctimes%204.1801%5Ctimes%20%28T_%7Bfinal%7D-83%29%5D)

Hence, the final temperature of the solution is 