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3 years ago

Hydrogen cyanide (HCN) is a product of combustion that:

Physics

1 answer:

bekas [8.4K]3 years ago

5 0

b. is equally as toxic as carbon monoxide (CO).

Explanation:

Hydrogen Cyanide is a product of combustion that is equally as toxic as carbon monoxide CO.

The gas is a product of combustion in blast furnances, coke ovens, e.t.c

Just like carbon monoxide, they are blood agents that are toxic in the body
Prolonged exposure to hydrogen cyanide can lead to death eventually.
The gas is a able to produce cyanide ion that seriously impacts the process of cellular respiration and bringing it to a halt.
It is listed as one of the dangerous chemical weapons.

Learn more:

Safety data sheets brainly.com/question/2188622

#learnwithBrainly

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A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara

otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

$r=2.3*sin70=2.161m$ (ball rotates in a circle)

If the system is in equilibrium, the tension is:

$Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}$

Replacing:

$\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}$

Replacing:

$v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s$

7 0

3 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$