Answer: c) with the same brightness
Explanation: The load in this case the bulb, is not polarized ( it has no positive and negative points) thus any connection relative to the battery (source) will have no effect on it brightness.
Also, brightness is a function of current and in this case the voltage ( from battery) and resistance of load (bulb) is constant, and according to ohms law (V=IR) if the current is constant at the first connection, it will be the same at the reversed connection.
The process is called neutralization.
Hope this helps!
Answer:
a) It is moving at
when reaches the ground.
b) It is moving at
when reaches the ground.
Explanation:
Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:
(1)
with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:
(2)
with m the mass and v the velocity.
Using (2) on (1):
(3)
In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):
(4)
Using (4) on (3):
(5)
That's the equation we're going to use on a) and b).
a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:


b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

Solving for initial velocity (when the boulder left the volcano):


Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2 (3 s.f)
Answer:
the signs of heat and work are; -Q and -W
Explanation:
The first law of thermodynamics is given by; ΔU = Q − W
where;
ΔU is the change in internal energy of a system,
Q is the net heat transfer (the sum of all heat transfer into and out of the system)
W is the net work done (the sum of all work done on or by the system).
Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).
Since work is done by the system, W remains negative.
Thus, the signs of heat and work are; -Q and - W