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nordsb [41]
3 years ago
9

In the pacific northwest orographic precipitation usually occurs where?

Physics
1 answer:
VMariaS [17]3 years ago
4 0
That’s the answer hope this helps!!!!

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Which of the following situations reflect the effects of deindividuation?
crimeas [40]

I think I’ve heard that in social studies if you do social studies today to the legislative branch of the racial equality because the legislative branch of the racial colony off the lot of slaves come out and carry Tubman and a lot of other slaves Rosa parks also
7 0
4 years ago
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To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Kryger [21]

Answer:

H = 10.05 m

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

so this is the speed at the top of flag pole

now we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

now the height of flag pole is given as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
3 years ago
I am starting a PowerPoint presentation about Cat-Eye Syndrome. I don't know how to start my sentence. How can I start?
ioda

Answer:

first introduce what is cat-eye syndrome e.g cat eye syndrome is....

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8 0
3 years ago
A) A 5.75 mL sample of mercury has a measured mass of 77.05 g. The density is ___________.
Kitty [74]

Answer:

Density of the sample will be 13.4 kg/L

Explanation:

We have given volume of the sample V=5.75mL=5.75\times 10^{-3}L

Mass of the sample M=77.05gram =77.05\times 10^{-3}kg

We have to find the density of the sample

Density of the sample is given by

Density=\frac{mass}{volume}=\frac{77.05\times 10^{-3}}{5.75\times 10^{-3}}=13.4kg/L

So density of the sample will be 13.4 kg/L

4 0
4 years ago
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

4 0
3 years ago
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