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4 years ago

What are some examples of vascular plants?

Physics

1 answer:

GarryVolchara [31]4 years ago

7 0

Hi Kayla

Examples are ;

1) Gymnosperms

2) Angiosperms

Hint; Gymnosperms create cones to house their seeds

Angiosperms; Create their seeds inside fruits

I hope that's help:0

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2. Light waves of the wavelength of 650 nm and 500 nm produce interference fringes on a screen at a distance of 1m from a double

satela [25.4K]

Answer:

least distance= 13mm

ratio of the lattice = 1 : 0.71 : 0.58

Explanation:

given λ₁ = 650nm = 650×10⁻⁹m, λ₂ = 500nm = 500×10⁻⁹m

5 0

3 years ago

I throw a ball upward at an initial speed of 20 m/s. How much time does it take before the ball slows to a speed of 0 m/s

masha68 [24]

20/9.8 = 2.0 seconds. The ball stops after 2 seconds.

5 0

3 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

2 years ago

How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10

Karo-lina-s [1.5K]

Answer:

$-2.00876\times 10^{-18}\ J$

Explanation:

v = Speed of electron = $2.1\times 10^6\ m/s$ (generally the order of magnitude is 6)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

Work done would be done by

$W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J$

The work required to stop the electron is $-2.00876\times 10^{-18}\ J$

6 0

4 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

$J=\frac{E}{\rho}$ (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

$J=\frac{I}{A}$ (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

$\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}$

Next, you replace for all variables:

$I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A$

hence, the current in the wire is 4.75A

4 0

3 years ago