All categories

3 years ago

Drosera magnifica is an organism that was first discovered in 2015.

Physics

1 answer:

mestny [16]3 years ago

3 0

Answer:

The correct answer is - Plantae.

Explanation:

Drosera magnifica is discovered in 2015 for the first time and the characteristics this organism's cell shows are -

- permanent vacuoles

- surrounded by cellulose layer

Vacuoles are present in both Plantae and Animalia kingdom of the eukaryotic organism but in animal cells, there are small and numerous vacuoles present and they are not permanent whereas in plant cells vacuoles are present permanently.

The cell of an animal cell has no surrounding layer other than cell membrane while in the plant cell there is a supporting and protecting layer of cellulose cell wall present.

On the basis of the given characteristics, it is confirmed that the Drosera magnifica belongs to Plantae kingdom.

You might be interested in

Intermolecular forces hold

blagie [28]

I believe Intermolecular forces hold, molecules, ions, and atoms? But I would see if that doesn't sound familiar check it with a site or something?

3 0

3 years ago

Define resistance and describe what would happen to a lightbulb if the voltage increase but the resistance stay the same?

artcher [175]

The first question is how much of a voltage increase are we looking at. If it has a 110 voltage rating and you put it across a 220 source, you will see one flash and then the bulb is no more. Nothing will revive it.

If it is rated at 110 and you put 130 across it, there's no problem but the bulb will burn out sooner than it would if you just put 110 across it.

So you raise the voltage and the resistance stays the same, the current will increase. That's why it will burn out sooner.

V = I * R

The equation is a direct variation. If the voltage goes up the current goes up. If the voltage goes down, the current goes down providing that the resistance stays the same in both cases.

The second question is what is resistance? Resistance in Electricity is the ability of an electric current to go in one direction freeing up as many electrons as it can. The MORE free electrons, the lower the resistance. The FEWER free electrons the higher the resistance.

Here' the kicker. Ready? More and Less are probably the two most important words in beginning science.

The More the resistance, the Less the current flow. That's a really important consideration in battery drain in a watch (or modern day calculator). The More the Battery Drain, the Less time it will last.

Always be careful when more and Less are around.

7 0

3 years ago

A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti

NeX [460]

Answer:

a) T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ , b) T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$ , d) m₂ = m₁ ( $\frac{ L}{2d} -1$ )

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

∑ τ = 0

T_A d - W₂ x -W₁ L/2 = 0

T_A = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$

b) the tension in string B

we write the expression of the translational equilibrium

∑ F = 0

T_A - W₂ - W₁ - T_B = 0

T_B = T_A -W₂ - W₁

T_ B = $\frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )$ - g m₂ - g m₁

T_B = g [m₂ ( $\frac{x}{d} -1$ ) + m₁ ( $\frac{L}{ 2d} -1$ ) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

g m₂ (d -x) - g m₁ (0.5 L -d) + 0 = 0

m₂ (d-x) = m₁ (0.5 L- d)

m₂ x = m₂ d - m₁ (0.5 L- d)

x = $d - \frac{m_1}{m_2} \ \frac{L}{2d}$

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

0 = d - \frac{m_1}{m_2} \ \frac{L}{2d}

$\frac{m_1}{m_2} \ (0.5L -d) = d$