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2 years ago

How are cactus adapted to survive in deserts?

Physics

1 answer:

andreev551 [17]2 years ago

7 0

1. They have evolved their leaves into spikes for minimum water loss through transpiration.

2. They have a waxy layer for minimum water loss.

3. They have thick walls for minimum water loss.

4. They can take water from atmosphere.

5. They change the photo energy from Sun into an intermediate stage and store it, so that they can make food even in night.

You might be interested in

True or False: In a parallel circuit, if one path is broken, electrons continue to flow to the

Snezhnost [94]

Answer:

false

Explanation:

this is what google said

No because the path the electricity needs to follow is broken. In a parallel circuit, electricity has more than one path to follow. Electrons can follow different paths as they flow from the negative side of the battery to the positive side.

4 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}

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

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

m₂ = m₁ $\frac{0.5 L -d}{d}$

m₂ = m₁ ( $\frac{ L}{2d} -1$ )

5 0

3 years ago

Suppose there are 100,000 signaling societies in our galaxy. What is the average distance between civilizations, assuming that c

Leokris [45]

Answer:

The average distance is 999.936 light years.

Solution:

As per the question:

Ours is the Milky Way Galaxy with the diameter:

D = 1,00,000 ly

Radius, R = $\frac{D}{2} = \frac{1,00,000}{2} = 50,000\ ly$

Now,

A = $\pi R^{2} = 50,000^{2}\pi = 7.8539\times 10^{9}\ ly^{2}$

The area to each civilization, A' is:

A' = $\frac{A}{10000} = \frac{7.8539\times 10^{9}}{10000} = 7.8539\times 10^{5}\ ly^{2}$

The radius of each area of civilization, r:

$A' = \pi r^{2}$

$r = \sqrt{\frac{A'}{\pi}}$

$r = \sqrt{\frac{7.8539\times 10^{5}}{\pi}} = 499.968\ ly$

Thus the average separation distance between civilizations is given as:

2r = $2\times 499.968 = 999.936\ ly$

8 0

3 years ago

What tissue composes the basement membrane surrounds cushions blood vessels and nerves

xxTIMURxx [149]

Answer:

Explanation:

Fascia is loose connective tissue that surrounds and interpenetrates all components of the human body including muscles, nerves, blood vessels, and organs. It provides structural integrity, serves as a matrix for intercellular communication, and is involved in biochemical and bioelectric signaling.

5 0

3 years ago

a box being pushed with a force of 85N right slides along the floor with a constant speed of 0.5m/s. What is the force of slidin

Sav [38]

Answer:

$85\; \rm N$ .

Explanation:

The box is sliding with a constant speed in a fixed direction (to the right.) In other words the velocity of this box is constant. Hence, this box would be in a translational equilibrium. The acceleration of this box would be zero.

By Newton's Second Law of motion, the net force on this box would be $0$ . In other words, forces on this box are balanced.

The question is asking for the size of the friction on the box. Assuming that the floor is horizontal. The friction on this box would also be horizontal,

The only other force that could balance that friction would be the $85\; \rm N$ push to the right. The direction of this push is horizontal (to the right.) Hence, the entirety of that $85\; \rm N\!$ would be in the horizontal direction.

Thus, forces on this box in the horizontal direction would be:

The $85\; \rm N$ push to the right.
Friction that opposes the rightward motion of the box (that is, to the left.)

Since these two forces must balance each other, the size of the friction would also be $85\; \rm N$ .

6 0

3 years ago