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MrRa [10]
3 years ago
15

How to determine the amount of resistance when two or more resistors are in parallel.

Physics
1 answer:
Elena L [17]3 years ago
3 0

Answer: 1/R = 1/R1 + 1/R2+ ...+ 1/Rn

R is resistance of system in which there are resistors R1, R2 , ... Rn parallel.

You might be interested in
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
BARSIC [14]

Answer:

The x component of the resultant force is -7.27N.

Explanation:

To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

|F_{1x}|=|F_1|\cos\theta_1=9.20N\cos62.0\°=4.31N \\|F_{2x}|=|F_2|\cos\theta_2=5.00N\cos53.6\°=2.96N

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

F_{1x}=-4.31N;\\F_{2x}=-2.96N

Finally, we sum both components to obtain the component of the resultant force:

F_{Rx}=-4.31N-2.96N=-7.27N

In words, the x component of the resultant force is -7.27N.

6 0
3 years ago
I need help thanks :)))))))))
dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end.  It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one.  It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0
3 years ago
true or false Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
salantis [7]
Hello Micu212006 


Question: <span> Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
</span><span>
Answer: True


Hope This Helps!
-Chris </span>
8 0
3 years ago
A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,
Nutka1998 [239]
Good morning.

We have:

\mathsf{\overset{\to}{a} = 29\overset{\to}{j}}

Where j is the unitary vector in the direction of the y-axis.

We have that 

\mathsf{\overset{\to}{a}+\overset{\to}{b} = -18\overset{\to}{j}}

We add the vector -a to both sides:

\mathsf{\overset{\to}{b} = -18\overset{\to}{j} -\overset{\to}{a} = -18\overset{\to}{j} -29\overset{\to}{j}}\\ \\ \mathsf{\overset{\to}{b}=-47\overset{\to}{j}}


Therefore, the magnitude of b is 47 units.
5 0
3 years ago
Read 2 more answers
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