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3 years ago

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Help plzzz itz importannnttt

2 answers:

krok68 [10]3 years ago

4 0

Answer:

➢ ➢ ➢ ✔3. How did Nazis treat their enemies?✔3. How did Nazis treat their enemies?

notka56 [123]3 years ago

3 0

I dont know the answer but heres some eagles for good luck

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The element selenium (Se) bonds with chlorine (Cl) to make the formula SeCl2 Chlorine is more electronegative than selenium. Wha

Akimi4 [234]

Answer:

Selenium dichloride

Explanation:

Selenium (Se) and Chlorine (Cl) are both elements capable of combining together to form a compound with the chemical formula; SeCl2. Since the chlorine atom is more electronegative than selenium atom, the chlorine pulls more electrons towards itself to form an IONIC bond.

The SeCl2 compound formed is called Selenium dichloride as two atoms of Chlorine are needed to combine with one atom of Selenium to form the compound.

7 0

3 years ago

a snail takes 16 minutes 40 seconds to cover a distance of 1 m calculate the average speed of the swimmer

matrenka [14]

The snail’s speed is 0.001042. Hope this helps!

8 0

3 years ago

Read 2 more answers

How many resistors are found in this circuit? <br> A) 0<br> B) 1<br> C) 2<br> D) 3

Dmitriy789 [7]

Show us the pictures I don't see it

6 0

3 years ago

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0

3 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

$= 343[\dfrac{508}{490}-1]$

$v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

$=\dfrac{12.6^2}{2\times 9.8}$

=8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

$= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

=0.293 m

total distance

= 8.1 + 0.293 = 8.392 m

3 0

3 years ago