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4 years ago

Why don't nonvascular plants grow very tall?

1 answer:

4 0

Because non vasculars essentially rely only on diffusion/osmosis for water transport, and getting water up very high with osmosis is extremely difficult.

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A wave has a wavelength of 10 mm and a frequency of 14 hertz. What is its speed?

Katena32 [7]

0.14 meters per second .

3 0

4 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm

Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]

here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]

= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0

3 years ago

An inductor is connected to a 26.5 Hz power supply that produces a 41.2 V rms voltage. What minimum inductance is needed to keep

alexira [117]

Answer:

The minimum inductance needed is 2.78 H

Explanation:

Given;

frequency of the AC, f = 26.5 Hz

the root mean square voltage in the circuit, $V_{rms}$ = 41.2 V

the maximum current in the circuit, I₀ = 126 mA

The root mean square current is given by;

$I_{rms} = \frac{I_o}{\sqrt{2} } \\\\I_{rms} = \frac{126*10^{-3}}{\sqrt{2} }\\\\I_{rms} =0.0891 \ A$

The inductive reactance is given by;

$X_l = \frac{V_{rms}}{I_{rms}} \\\\X_l= \frac{41.2}{0.0891}\\\\X_l = 462.4 \ ohms$

The minimum inductance needed is given by;

$X_l = \omega L\\\\X_l = 2\pi fL\\\\L = \frac{X_l}{2\pi f}\\\\L = \frac{462.4}{2\pi *26.5}\\\\L = 2.78 \ H$

Therefore, the minimum inductance needed is 2.78 H

7 0

3 years ago

Which event is an example of condensation?

ale4655 [162]

Answer: D

If the fog disappears when the Sun comes out, then this is an example of condensation because:

the Sun actually dries up the fog, and it makes it into higher clouds.

Hope this helps you!

3 0

3 years ago