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2 years ago

5

How can you tell the stigma from the stamens in the flowers

1 answer:

pishuonlain [190]2 years ago

6 0

Answer:

for wind pollinated flowers they have long feathery stigma that hangs outside the flower while insect pollinated have short stigma that is shorter than even the petals and it's inside the flower

Also thestigma has something that looks like a bulb underneath it and that is the ovary while the stamen is just straight

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Can you make sunglasses from new 3D-glasses???

katen-ka-za [31]

Yes you can make sunglasses from 3d glasses

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2 years ago

Read 2 more answers

Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips

Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

$f = \frac{v}{\lambda}$

Then the relation between two different frequencies with same wavelength would be

$\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}$

$\frac{f'}{f} = \frac{v'}{v}$

The beat frequency heard when the two strings are sounded simultaneously is

$f_{beat} = f-f'$

$f_{beat} = f(1-\frac{f'}{f})$

$f_{beat} = f(1-\frac{v'}{v})$

We have the velocity of the transverse waves in stretched string as

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{200N}{\mu}}$

And,

$v' = \sqrt{\frac{196N}{\mu}}$

Therefore the relation between the two is,

$\frac{v'}{v} = \sqrt{\frac{192}{200}}$

$\frac{v'}{v} = \sqrt{0.96}$

Finally substituting this value at the frequency beat equation we have

$f_{beat} = 590(1-\sqrt{0-96})$

$f_{beat} = 11.92Hz$

Therefore the beats per second are 11.92Hz

4 0

2 years ago

What are the relative ages of the features in order of oldest to youngest?

yuradex [85]

Answer:

Layer 1, Rock 2, Rock 1, Fault

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2 years ago

A seesaw pivots as shown in below. What is the net torque about the pivot point?

rodikova [14]

Answer:

2.3 Nm clockwise

Explanation:

Take counterclockwise to be positive and clockwise to be negative.

∑τ = (3 N) (2.5 m) − (7 N) (1.4 m)

∑τ = 7.5 Nm − 9.8 Nm

∑τ = -2.3 Nm

The net torque is 2.3 Nm clockwise.

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3 years ago

1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame

IgorC [24]

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

4 0

2 years ago