When a single high-resistance (long) bulb is connected to a 1.5 V battery, the current through the battery is about 80 mA. If you add another high-resistance (long) bulb in parallel, the battery current of course increases to 160 mA. Select all of the true statements given this situation. a. The battery is ohmic.
b. The battery is not ohmic.
c. Current through the battery is proportional to ΔV across the battery.
d. Current through the battery is not proportional to ΔV across the battery.
e. The battery always puts out the same current.

a) True. The battery obeys ohm's law, it is formed by an ideal source with a fixed internal resistance

c) True. Ohm's law is V = iR therefore voltage and current are proportional

Explanation:

In this problem let's analyze the load of the system, when a resistance is placed the current is 80 mA, if we place two resistors in parallel the voltage remains the same, but the current is divided between each resistance (bulb), therefore the current in the battery it must be 160 mA

Let's analyze the answers

a) True. The battery obeys ohm's law, it is formed by an ideal source with a fixed internal resistance

b) false

c) True. Ohm's law is V = iR therefore voltage and current are proportional

d) False

e) False. The current coming out of the battery is proportional to the load placed

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015