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marysya [2.9K]
2 years ago
10

Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

ojection Pa of R onto axi

Physics
1 answer:
Liula [17]2 years ago
6 0

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}  

therefore : Ra = \frac{sin120^o * 750}{sin 35^o} = 1132 N

To determine the scalar component Rb

\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}

therefore : Rb = \frac{sin 25^o * 750}{sin 35^o}  = 522.6 N

To determine the orthogonal projection Pa of R onto

Pa = 750 cos25^o = 679.7 N

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Two parallel wires carry a current I in the same direction. Midway between these wires is a third wire, also parallel to the oth
Luda [366]

Answer:

Force is repulsive hence direction of force is away from wire

Explanation:

The first thing will be to draw a figure showing the condition,

Lets takeI attractive force as +ve and repulsive force as - ve and thereafter calculating net force on outer left wire due to other wires, net force comes out to be - ve which tells us that force is repulsive, hence direction of force is away from wire as shown in figure in the attachment.

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2 years ago
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The temperature and pressure at the surface of Mars during a Martian spring day were determined to be -50 °C and 900 Pa, respect
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Answer:

T = 273 + (-50) = 273 – 50 = 223 K

R = 188.82 J / kg K for CO2

Density (Martian Atmosphere) = P / RT = 900 / 188.92 x 223 = 900 / 42129.16 = 0.0213 kg / m^{3}

T = 273 +18 = 291 K, R = 287 J / kg k (for air) P = 101.6 k Pa = 101600 Pa

Density (Earth Atmosphere) = P / RT = 101600 / 287 x 291 = 1.216 kg / m^{3}

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2 years ago
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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

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Use the periodic table and your knowledge of isotopes to complete these statements.
klasskru [66]
<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.

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<span><span>Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of </span><span> ⇒ 18</span>.</span>

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