Answer:
Explanation:
Given:
Initial θ = 0 rad (from rest)
Final θ = 14.3 rad
Time, t = 5 s
B.
Angular velocity, w = dθ / dt
= (14.3 - 0)/5
= 2.86 rad/s
A.
Acceleration, ao = dw/dt
Initial angular velocity, wi = 0 rad/s (from rest)
Final angular velocity, wf = 2.86 rad/s
a = (2.86 - 0)/5
= 0.572 rad/s^2
The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.
The current flowing through the lamp is given by Ohm's law:

where
V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.
The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.
<h2>
Answer:
</h2>
The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting
the radius of the orbit (measured from the center of the planet to the satellite).
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>
In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.
Answer:
351 ohm
720 ohm
Explanation:
When c and d are open:
Terminals c and d are open.. If you redraw the circuit as below, you can see that the two resistors in the first column are in parallel as, they are connected together at both pairs of terminals (due to the short).
Hence, we have a pair of parallel resistors:
Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms
Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms
Now these two sets are in series with another Hence,
Req = Req1 + Req2 = 216 + 135 = 351 ohms
Answer: 351 ohms
When c and d are shorted:
The current will flow through the least resistant path naturally from resistors R3 and R1 or R4.
Both of these resistor lie in a single path placing the resistors in series to one another, hence
Req = R3 + R1 = 180 + 540 = 720 ohms
Answer:720 ohms
Answer:
The current drawn by Horace’s reading glasses is 0.8 A.
Explanation:
Given that,
Resistance of each bulb, R = 2 ohms
Voltage of the system, V = 3.2 volts
These two bulbs are connected in series. The equivalent resistance will be 2 ohms +2 ohms = 4 ohms
Let I is the current drawn by Horace’s reading glasses. Using Ohm's law to find it such that :

So, the current drawn by Horace’s reading glasses is 0.8 A.