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stiks02 [169]
3 years ago
7

a reaction occurs in which carbon combines with sulfur to form carbon disulfide is this a chemical reaction or a nuclear reactio

n and how do you know
Physics
2 answers:
romanna [79]3 years ago
7 0

Answer:

a chemical reaction where electrons mix.

Explanation:

A nuclear reaction is when atomic nuclei combine, forming a new chemical element

A chemical reaction occurs when electrons from the last layer of atoms mix, forming molecules that have the same chemical elements as the reactants.

According to these definitions

           C + 2 S → CS2

in a chemical reaction where electrons mix.

podryga [215]3 years ago
6 0

Answer:

This is a chemical reaction, because only the electrons were rearranged.

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Which graph would you use to show the percentage of cookies sold by each
tatuchka [14]

Answer:

You would use a bar graph

4 0
3 years ago
The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.
andrew-mc [135]

Answer:

answer is 3.05v

Explanation:

hope it is helpful and briliant

6 0
2 years ago
A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = I_t

Rotational inertia (I_r) of the record = 0.61\times I_t

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as  \omega _i\  ,\  \omega_f respectively.

Note :

Angular momentum (L) = Product of moment of inertia  (I)  and angular velocity (\omega) .  

Lets say,

⇒ initial angular momentum = final angular momentum

⇒  L_i=L_f

⇒ (I_t)\times \omega_i = (I_t+I_r)\times \omega_f

⇒ \omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i) ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ K_i=\frac{I_t\times \omega_i^2}{2}       ⇒ K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}

Their ratios:

⇒ \frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }    

⇒ \frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}

Plugging the values of  \omega _f^2 as \omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2 from equation (i) in the ratios of the Kinetic energies.

⇒ \frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}

Now,

The Kinetic energy lost in fraction can be written as:

⇒ \frac{K_f-K_i}{K_i}

Now re-arranging the terms.

\frac{K_f-K_i}{K_i}  =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}

Plugging the values of  I_r and I_t .

⇒ \frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788

To find the percentage we have to multiply it with 100 and here negative means for loss of Kinetic energy.

⇒ \frac{K_f}{K_i} = =-0.3788\times 100= 37.88

So the percentage of the initial Kinetic energy lost is 37.88

4 0
2 years ago
The current in a toaster is 20 amps, and its voltage is 200 volts. What is the resistance in the toaster?
alukav5142 [94]
<span>Ohm's law deals with the relation between voltage and current in an ideal conductor. It states that: Potential difference across a conductor is proportional to the current that pass through it. It is expressed as V=IR. 

V = IR
200 = 20R
R = 10 ohms</span>
8 0
2 years ago
Read 2 more answers
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

4 0
2 years ago
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