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3 years ago

Which are base units used in the metric system? Check all that apply.

Physics

2 answers:

Anestetic [448]3 years ago

8 0

Quarts - X

<h3>Liters - √</h3><h3>Grams - √</h3>

Degrees Fahrenheit - X

<h3>Degrees Celsius - √</h3>

Pints - X

I've tried my best with this, sorry if any are wrong.

Hope this helps,

Davinia.

Shtirlitz [24]3 years ago

7 0

Liters

Grams

Degrees Celsius

The other answer choices are from the imperial system

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How is the L-section filter useful at both low and high frequencies?

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Answer: High pass filters, and in particular LC high pass filters are used in many RF applications where they block the lower frequencies and allow through higher frequency signals. Typically LC filters are used for the higher radio frequencies where active filters are not so manageable, and inductors more appropriate.

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2 years ago

Read 2 more answers

Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

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3 years ago

You're driving down the highway late one night at 21 m/s when a deer steps onto the road 35 m in front of you. Your reaction tim

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Speed with which initially car is moving is 21 m/s

Reaction time = 0.50 s

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$d = 22.05 m$

so total distance moved before it stop

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so the distance from deer is 35 - 32.55 = 2.45 m

now to find the maximum speed with we can move we will assume that we will just touch the deer when we stop

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$v_f^2 - v_i^2 = 2 ad$

$0 - v^2 = 2(-10)(24.5)$

$v = 22.1 m/s$

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