What are the characteristics of prokaryotes?

1 answer:

5 0

Prokaryotes lack an organized nucleus and other membrane-bound organelles. Prokaryotic DNA is found in a central part of the cell called the nucleoid. The cell wall of a prokaryote acts as an extra layer of protection helps maintain cell shape, and prevents dehydration.

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A 1056-hertz tuning fork is sounded at the same time a piano note is struck. You hear three beats per second. What is the freque

boyakko [2]

Answer:

Explanation:

There is not enough information in the question. A beat is the observed differences between two frequencies.

Hence, 3 beat represents a difference of 3-hertz between the two frequencies. So the frequency of the piano string could be 1056+3 = 1059-hertz or 1056-3= 1053-hertz.

In order to ascertain the actual value i.e if it is an increase or a decrease in frequency observed, we need more information.

4 0

3 years ago

7) A kite 50 ft above the ground moves horizontally at a speed of 4 ft/sec. At what rate is the length of the string increasing

d1i1m1o1n [39]

Imagine a right triangle where the legs represent the horizontal and vertical lengths of the string and the hypotenuse represents the length of the string.

Let us assign some values:

x = horizontal length in feet

50 = vertical length in feet

L = length of the string in feet

Because we are modeling these quantities with a right triangle, we can use the Pythagorean theorem to relate them with the following equation:

L² = x² + 50²

We want to find an equation for the change of L over time, so first differentiate both sides with respect to time t then solve for dL/dt:

2L(dL/dt) = 2x(dx/dt)

dL/dt = (x/L)(dx/dt)

First let's solve for x at the moment in time described in the problem using the Pythagorean theorem:

L² = x² + 50²

Given values:

L = 100ft

Plug in and solve for x:

100² = x² + 50²

x = 86.6ft

Now let's find dL/dt. Given values:

x = 86.6ft, L = 100ft, dx/dt = 4ft/sec

Plug in and solve for dL/dt:

dL/dt = (86.6/100)(4)

dL/dt = 3.46ft/sec

3 0

4 years ago

A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a

Schach [20]

Before it hits the sand bed, the meteorite is accelerating uniformly with $g=9.80\dfrac{\rm m}{\mathrm s^2}$ , so that its speed $v$ satisfies

$v^2-{v_0}^2=-2g\Delta y$

where $v_0=90.0\dfrac{\rm m}{\rm s}$ is its initial speed and $\Delta y=(0-850)\,\mathrm{km}=-850\,\mathrm{km}$ is its change in altitude. Notice that we're taking the meteorite's starting position in the atmosphere to be the origin, and the downward direction to be negative. Now,

$v^2-\left(-90.0\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(-850,000\,\mathrm m)\implies\boxed{v=4080\dfrac{\rm m}{\rm s}}$