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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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Which term refers to the energy that is stored or released during a change of state of water

Dima020 [189]

The appropriate term is latent heat. This energy is released as the water changes state from a gas to liquid....a liquid to solid etc. the latent heat is either absorbed or given off by the water as it changes its physical state. Latent heat of fusion is associated with freezing a liquid or melting a solid.

8 0

3 years ago

Read 2 more answers

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

3 years ago

Which of the following is an example of Class 3 lever system

insens350 [35]

If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.

6 0

3 years ago

Read 2 more answers

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a

marishachu [46]

Answer:

-9.46 N

Explanation:

In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.

This value will be the projection of the force vector, on the x-axis.

For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:

cos θ = $\frac{x}{r}$

In this case, x = F₁ₓ, and r = F₁

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º-55º = 125º

⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N

We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ = $\frac{x}{r}$

In this case, x = F₂ₓ, and r = F₂

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º + 53.3º = 233.3º

⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N

The total component of both forces along the x axis, can be found just adding both components, as follows:

Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N

5 0

3 years ago