All categories

3 years ago

According to the dichotomous key, stentor would be a member of the kingdom

Physics

1 answer:

dalvyx [7]3 years ago

5 0

D) Protista

Explanation:

The Dichotomous key is a organized set of couplets of mutual characteristics of biological organisms. This key begins with general characteristics and leads couplets indicating specific characteristics.
Dichotomous key is a key that allow the user to identify type of items in natural world. Stentor is one of the largest protozoa found in water.
Stentor is a single cell. An organism can be of 2 mm in length making them visible to human eye and it is even larger than many multi celled organisms. Stentor is eukaryote means it has nuclear membrane. It is commonly found in fresh water ponds.

You might be interested in

Need help ASAP please Thanks

Karo-lina-s [1.5K]

Answer: D
Explanation: there is less light at that point.

3 0

3 years ago

Explain what is happening in this picture

Assoli18 [71]

Answer:

in this video waves are coming up for the BOTTOM to the top of the sandbar

7 0

3 years ago

suppose you have a 69.0-kg wooden crate resting on a wood floor. what maximum force can you exert horizontally on the crate with

marusya05 [52]

You've got a 69.0-kg wooden crate on a wooden floor. The box can withstand a force of up to 338N in a horizontal direction without being moved. Following this, the wooden creates moving stats.

In order to calculate the friction coefficient, divide the force pushing two objects together by the force acting between them. friction coefficient might be 0 or one. They can be split into two categories: friction coefficient that is static. Kinetic friction coefficient (also known as sliding coefficient of friction).

the acceleration brought on by the gravitational pull of large masses generally, gravitational , often known as the acceleration brought on by the Earth's gravitational pull and centrifugal force,

F= friction coefficient *M*g

F= 0.5*69*9.8

F=338N

Learn more about gravitational here

brainly.com/question/3009841

#SPJ4

7 0

1 year ago

Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.

jek_recluse [69]

According to the statement " Collision between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "

4 0

3 years ago

Read 2 more answers

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,