<span>temperature increases and molecular motion increases while shape becomes less defined.
</span>
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
Answer:
1. 0.45 s.
2. 4.41 m/s
Explanation:
From the question given above, the following data were obtained:
Height (h) = 1 m
Time (t) =?
Velocity (v) =?
1. Determination of the time taken for the pencil to hit the floor.
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.8
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s.
Thus, it will take 0.45 s for the pencil to hit the floor.
2. Determination of the velocity with which the pencil hit the floor.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 0.45 s.
Final velocity (v) =?
v = u + gt
v = 0 + (9.8 × 0.45)
v = 0 + 4.41
v = 4.41 m/s
Thus, the pencil hit the floor with a velocity of 4.41 m/s
Answer:
The speed of the heavier fragment is 0.335c.
Explanation:
Given that,
Mass of the lighter fragment 
Mass of the heavier fragment 
Speed of lighter fragment = 0.893c
We need to calculate the speed of the heavier fragment
Let v is the speed of the second fragment after decay
Using conservation of relativistic momentum













Hence, The speed of the heavier fragment is 0.335c.