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blondinia [14]
2 years ago
5

What structure holds donor sperm in earthworms

Physics
1 answer:
lidiya [134]2 years ago
4 0

Answer:

Explanation: As the earthworm passes through the slime tube, the tube will pass over the female pore picking up eggs. The tube will continue to move down the earthworm and pass over the male pore called the spermatheca which has the stored sperm called the spermatozoa.

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Suppose that the collector is held at a small negative voltage with respect to the grid. Will the accelerated electrons reach th
Leona [35]

Answer:

B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.

Explanation:

In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential

so according to the given situation, the option b is correct

And, the rest of the options are wrong

3 0
3 years ago
Which is more useful for storing thermal energy, the block of lead or the water in the calorimeter?
leonid [27]

I believe it is the block of lead

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3 years ago
Does anyone know the answer to all of these questions
Rasek [7]

Answer:

i don't understand the hw

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Q.1 Newton's second law gives the measure of--------------. 1)Acceleration 2) Force 3) Momentum 4)Angular momentum​
posledela

Answer:

1. Acceleration

Explanation:

Newtons Second law gives the measure of acceleration

4 0
3 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
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