1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].
2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].
3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].
4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].
Answer:
The net force acting on this object is 180.89 N.
Explanation:
Given that,
Mass = 3.00 kg
Coordinate of position of 
Coordinate of position of 
Time = 2.00 s
We need to calculate the acceleration

For x coordinates

On differentiate w.r.to t

On differentiate again w.r.to t

The acceleration in x axis at 2 sec

For y coordinates

On differentiate w.r.to t

On differentiate again w.r.to t

The acceleration in y axis at 2 sec

The acceleration is

We need to calculate the net force



The magnitude of the force


Hence, The net force acting on this object is 180.89 N.
Answer:
Option C
Maximum potential energy is at point R.
Explanation:
Potential energy is a product of mass, acceleration due to gravity and height ie
PE=mgh where PE is the potential energy, m is mass of an object, g is acceleration due to gravity whose value is normally taken as 9.81 and h is height. Since at point R we have the maximum height, the potential energy will be highest at this point.
Answer:
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By using ramps you can easily push or pull the object up the ramp.