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Stels [109]
3 years ago
11

Describe binary fission with amoeba.​

Physics
1 answer:
kirza4 [7]3 years ago
6 0
Amoeba reproduce by the common asexual reproduction method called binary fission
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What can happen to a photon of light at each change of medium?
Kryger [21]
Give me some answer choices and i will be happy to help
5 0
3 years ago
Olivia places her pet frog on a line to observe the frog motion. The line is divided into sections that measure 1 centimeter eac
Goshia [24]
There is no displacement. The frog is back where it began.
4 0
4 years ago
Read 2 more answers
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
What is the measure of a change in an object’s velocity?
tiny-mole [99]
It's a measure of the acceleration
6 0
3 years ago
la potencia mecánica, si Carlos realiza un trabajo de 95 J al subir un objeto del suelo a la mesa en un tiempo de 9 segundos.
irga5000 [103]

Answer:

P = 10,56 Watts

Explanation:

La formula de la potencia es:

P = W/t

donde P es la potencia, W el trabajo realizado y t el tiempo transcurrido.

Reemplazando los valores dados obtengo:

P = 95 J / 9 seg

P = 10,56 Watts

4 0
4 years ago
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