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3 years ago

A(n) _______ is a pure substance that can't be broken down into simpler substances by chemical or physical means. A. element

1 answer:

3 0

An element is a pure substance that cant be broken down into simpler substances by chemical or physical means. hope it helps :)

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An alternating current is supplied to an electronic component with a rating that the voltage across it can never, even for an in

Vsevolod [243]

Answer:

A) $V_{rms}=8\sqrt{2} V$

Explanation:

Maximum voltage = $V_{max}=16 V$

Maximum voltage and rms voltage are related to each other by

$V_{max}=V_{rms} \times \sqrt{2} \\V_{rms}=\frac{V_{max}}{ \sqrt{2}}\\V_{rms}=\frac{16}{\sqrt{2}} \\V_{rms}=8\sqrt{2} V$

7 0

3 years ago

Wire A carries 4 A into a junction, wire B carries 5 A into the same junction, and another wire is connected to the junction. Wh

nydimaria [60]

Answer:9A

Explanation:

let the last wire be wire C

According to Kirchhoff's rule

the sum of all currents entering a junction must be equal to the sum of all currents leaving a junction

Ic=Ia+Ib

Ic= 4+5

Ic=9A

7 0

3 years ago

Which planet is most likely to have acid rain? A. Mercury B. Venus C. Mars D. Uranus

timofeeve [1]

Its VENUS because the planet is basicily a hot planet

3 0

3 years ago

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, $P_{gauge \ pressure}$ = 2.95 atm

Atmospheric pressure, $P_{atm}$ = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2$

where;

P₁ = $P_{gauge \ pressure} + P_{atm \ pressure}$

P₂ = $P_{atm}$

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

$P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }$

where;

$\rho$ is the density of seawater = 1030 kg/m³

$v_2 = \sqrt{ \frac{2(2.95*101325 \ + \ 1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s$

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0

3 years ago

If your weight is 120 pounds and your mass is 54 kilograms how would those values change if you were on the moon

BlackZzzverrR [31]

The gravitation acceleration on the moon is different than on Earth. It is 1.6 m/s^2. If you weigh 120 lbs, then you would multiply 120 pounds by the gravitational acceleration on the moon and then divide by the acceleration on Earth.

(120 lbs * 1.6) / 9.8 = 20 pounds.

The mass will always be the same no matter what planet you’re on, so it’s still 54 kg.

4 0

3 years ago