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lesya [120]
3 years ago
9

which element is used as rat poison and is found in the title of the murder mystery _______ and old lace

Physics
2 answers:
defon3 years ago
6 0

Explanation :

Arsenic and old lace is used in rat poison and is found in the title of the murder mystery.

Arsenic is a poisonous element. It is a metalloid that has some properties of metals and some of the nonmetals.

As it is a toxic thing. It can attack the digestive system, nervous system and on the rat's heart. If it is consumed by rats or humans in excess amount, the rat can kill very quickly.

So, Arsenic and old lace is the element.

alisha [4.7K]3 years ago
3 0
Arsenic and old lace is the title you are looking for

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Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 19.0 cmcm and carries
Lemur [1.5K]

Answer:

Explanation:

magnetic field due to circular wire

= μ₀ i / 2r

i is current and r is radius of coil .

Magnetic fields due to inner coil

μ₀ x 20 / (2 x 9.5 x 10⁻²)

Magnetic field due to outer coil

= μ₀ x I / (2 x 19 x 10⁻²) , I is the current to be calculated

Total field

μ₀ x 20 /( 2 x 9.5 x 10⁻²) +μ₀ x I / (2 x 19 x 10⁻²)  = 0

20 + I /2 = 0

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Where is the most of the liquid freshwater on earth located?
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A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What hap
Olenka [21]

Answer:

<em>His angular velocity will increase.</em>

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = I'ω'

where

I' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = mr'^{2}

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to I.

From

I'ω' = Iω

since I is now reduced, ω will be greater than ω'

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5 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

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