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Nutka1998 [239]

3 years ago

6

Составь словарный диктант из глаголов неопределённой формы которые есть в орфографическом словаре учебника рядом Перепишите слов

а в форме прошедшего времени множественного числа

1 answer:

Lilit [14]3 years ago

7 0

I cant understand tht wats the translation at

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A balloon is ascending at 12.4m/s at a height of 81.3m above the ground when a package is dropped. a) How long did it take to re

tankabanditka [31]

Answer:

3secs

Explanation:

Given the following parameters

height H= 81.3m

Velocity v = 12.4m/s

Required

Time it take to reach the ground

Using the equation of motion

H = ut+1/2gt²

81.3 = 12.4t + 1/2(9.8)t²

81.3 = 12.4t + 4.9t²

4.9t² + 12.4t - 81.3 = 0

Using the general formula to find t

t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)

t = -12.4±√153.76+1593.48/2(4.9)

t = -12.4±√1747.24/9.8

t = -12.4+41.8/9.8

t = 29.4/9.8

t = 3secs

Hence it took 3secs to reach the ground

5 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

25.16 what is the reaction product of acetic acid and ethylamine at room temperature?

Ne4ueva [31]

The ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature

<h3 /><h3>What is acetic acid ?</h3>

Acetic acid is a monofunctional carboxylic acid containing two carbon atoms. It acts as a protein solvent, food acidity regulator, antibacterial food preservative. It is a conjugate acid of an acetate.

Acetic acid is used in the production of acetic anhydride, cellulose acetate, vinyl acetate monomer, acetic ester, chloroacetic acid, plastics, dyes, insecticides, photographic chemicals, and rubber. Other commercial uses include the production of vitamins, antibiotics, hormones, organic chemicals, and as a food additive. Typical concentrations of acetic acid found naturally in foods are 700 to 1200 milligrams/kg (mg/kg) in wine, up to 860 mg/kg in aged cheeses, and 2.8 mg/kg in aged cheeses. fresh orange juice.

learn more about acetic acid, visit;

brainly.com/question/16970860

#SPJ4

7 0

1 year ago

Which protects a cell and forms it’s boundary

ivann1987 [24]

Answer:

For a plant cell: The cell wall and the cell membrane

For animal cells: Just the cell membrane

5 0

3 years ago

Waves in a fish bowl jostled by the Thingamajigger move to the sides at an average velocity of 0.50 m/s. If they occur once ever

Colt1911 [192]

Answer:

0.125 m

Explanation:

In this problem, we have:

v = 0.50 m/s is the average velocity of the wave

T = 0.25 s is the period of the wave

We can find the frequency of the wave, which is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.25 s}=4 Hz$

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

$\lambda=\frac{v}{f}$

Substituting the numerical values, we find

$\lambda=\frac{0.5 m/s}{4 Hz}=0.125 m$

4 0

3 years ago