<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 
When Venus put in a different battery with higher voltage ... no matter
what other components were in the circuit ... the voltage across the
light bulb, and the current through it, both had to increase, and the
light bulb had to shine brighter than before.
Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
Answer:
Explanation:
Given
radius r=2.96 mm
Tension T=2.4 N
time taken=0.74 s
Let
be the angular acceleration







Angular momentum




Answer:
The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.
Explanation:
Given;
emf of the battery, V = 12 V
resistance of the resistor, R = 100-Ω
time of current flow, t = 1 min
charge of 1 electron = 1.602 x 10¹⁹ C
The current through this circuit is given by;
I = V / R
I = (12) / (100)
I = 0.12 A
The quantity of charge or electron flowing the wire in the given time is calculated as;
Q =It
where;
I is the current flowing through the wire
t is the time of current flow = 1 x 60s = 60 s
Q = 0.12 x 60
Q = 7.2 C
1.602 x 10⁻¹⁹ C --------------- 1 electron
7.2 C -----------------------------? electron

Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.