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3 years ago

What is the reason so many poisonings go undetected at first?

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Poisonings mostly have no taste or smell. Thallium is one of the poisons that are hard to detect. Most of these poisons are always prescribed to many patients and can be misused. We also have to take into consideration thay poisoning attacks our immune system and internal organs. Side effects of poisoning are flue like. So when the individual is feeling ill. nobody is going to think it’s poison at first until about 30 mins to an hour. that’s why it takes so long to be undetected at first. because of the symptoms, smell, taste..... it can really be easy to poison someone

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zavuch27 [327]

Explanation:

It is given that,

Diameter of the peach pie, d = 9 inches

Radius of the pie, r = 4.5 inches

The tray is rotated such that the rim of the pie plate moves through a distance of 183 inches, d = 183 inches

Let $\theta$ is the angular distance that the pie plate has moved through.

It is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{183}{4.5}$

$\theta=40.66\ radian$

Since, 1 radian = 57.29 degrees

$\theta=2329.64\ degrees$

Since, 1 radian = 0.159155 revolution

$\theta=6.47124\ revolution$

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3 years ago

You push against a steamer trunk with a force of 800 n at an angle alpha with the horizontal . the trunk is on a flat floor and

galina1969 [7]

The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
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3 years ago

I NEED HELP PLEASE, THANKS! :)

makkiz [27]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference

EastWind [94]

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

We know that

Electric potential difference ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

$\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J$

ΔU =18.79 mJ

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