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GREYUIT [131]
3 years ago
7

What protists do not have definite shape

Physics
1 answer:
enyata [817]3 years ago
6 0

Amoeba

Explanation:A type of protozoan that does not have a definite shape. It moves by using pseudo pods (false feet).

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The series in the He spectrum that corresponds to the set of transitions where the electron falls from a higher level to the nf
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Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:  
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>

<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>

<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>

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A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0
2 years ago
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