Answer: A voltmeter must have a high resistance where as an ammeter must have a low resistance.
Explanation:
A voltmeter is a device which is connected in parallel to the component across which voltage needs to be measured. In a parallel circuit voltage drop is same at the nodes. The parallel connection must not offer easier path for current to divert from the main circuit and travel. Thus, a voltmeter must have high resistance.
On the other hand, an ammeter which is used to measure current in the circuit must have low resistance as it is connected in series. It should not offer resistance as it would reduce the actual current and measurement would be inaccurate.
Answer:
there is the increase the temperature of cold body and decrease the temperature of hot body
I believe the answer is A
Answer:
The horizontal distance is 0.64 m.
Explanation:
Initial velocity, u =2.5m/s
The maximum horizontal distance is

Answer:
each resistor is 540 Ω
Explanation:
Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance
defined by the formula:

Therefore, R/3 is the equivalent resistance of the initial circuit.
In the second circuit, two of the resistors are in parallel, so they are equivalent to:

and when this is combined with the third resistor in series, the equivalent resistance (
) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

The problem states that the difference between the equivalent resistances in both circuits is given by:

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:
