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Reil [10]
3 years ago
13

Match the following vocabulary terms to their definitions. 1. density a proposed explanation for a scientific problem 2. matter

standard metric unit of mass 3. control standard metric unit of length 4. meter anything that has mass and occupies space 5. hypothesis the mass of a substance per unit volume 6. kilogram a quantity in an experiment that remains unchanged or constant
Physics
2 answers:
zloy xaker [14]3 years ago
7 0
Density - the mass of a substance per unit volume
Matter - anything that has mass and occupies space
Control standard - a quantity that remains unchanged
Meter - metric unit of length
hypothesis - a proposed explanation for a scientific problem
kilogram - metric unit of mass
maw [93]3 years ago
6 0
<span>a proposed explanation for a scientific problem: hypothesis
</span><span>standard metric unit of mass: kilogram
</span><span>standard metric unit of length: meter
</span><span>anything that has mass and occupies space: matter
</span><span>the mass of a substance per unit volume: density
</span><span>a quantity in an experiment that remains unchanged or constant: control</span>
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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
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Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0
3 years ago
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