What do you thing about Dr.Mahathir(new president of Malaysian)

On a hot day, warm air on land rises and collides with cool air. Which of these weather conditions is most likely to follow?

Sliva [168]

I think its......C. Hurricane

3 0

3 years ago

Read 2 more answers

Find the value of currents through each branch

Irina-Kira [14]

Answer:

the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

(I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

$\left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]$

These equations have the solution ...

$\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]$

This means the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

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It helps to be familiar with the formulas for resistors in series and parallel.

8 0

3 years ago

What factor determines an object's momentum?

vlada-n [284]

Momentum = mass x velocity

So both mass and velocity affect an object's momentum.

3 0

3 years ago

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

$T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s$

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m$

From Hooke's law

$mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2$

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0

3 years ago

A person using an oxygen mask is breathing air that is 33% oxygen. what is the partial pressure of the o2, when the air pressure

yawa3891 [41]

The partial pressure of the O2 is 36.3 kiloPascal when the air pressure in the mask is 110 kiloPascal based on the isotherm relation. This problem can be solved by using the isotherm relation equation which stated as Vx/Vtot = px/ptot, where V represents volume, p represents the pressure, x represents the partial gas, and tot represents the total gas<span>. Calculation: 33/100 = px/110 --> px = 36.3</span>

5 0

3 years ago