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zhannawk [14.2K]
4 years ago
15

What is the net force of this object ?22newtons36 Newtons0 newtons8 newtons

Physics
2 answers:
antiseptic1488 [7]4 years ago
4 0
There is no net vertical force.
The 20N upward is exactly balanced by the 20N downward.

There IS a net horizontal force.
(22N to the right) + (14N to the left)  =  8 N to the right.

The net force acting on this object is 8 newtons to the right. 
Natali5045456 [20]4 years ago
3 0

The answer is: 8 Newtons.

I got 100% on the test.

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Answer:

Approximately 111\; {\rm \Omega}.

Explanation:

It is given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega} are connected in a circuit in parallel.

Assume that this circuit is powered with a direct current power supply of voltage V.

Since R_{1} and R_{2} are connected in parallel, the voltage across the two resistors would both be V. Thus, the current going through the two resistors would be (V / R_{1}) and (V / R_{2}), respectively.

Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: I = (V / R_{1}) + (V / R_{2}).

In other words, if the voltage across this circuit is V, the total current in this circuit would be I = (V / R_{1}) + (V / R_{2}). The (equivalent) resistance R of this circuit would be:

\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}.

Given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega}:

\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}.

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