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Sati [7]
3 years ago
9

2 bulbs in a series what is the current​

Physics
1 answer:
Basile [38]3 years ago
5 0
It doubles I believe
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If there is an object in top of another object, why does the upper object exert a downward normal force?​
rusak2 [61]

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.  

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

4 0
3 years ago
PLEASE HELP ASAP!! THIS IS DUE IN 15 MUNITES!! Explain what a Precipitate (Not Precipitation) is and tell me if it is a Physical
Amiraneli [1.4K]

Answer:

A precipitate is a solid that separates out of a liquid solution when it is supersaturated (insoluble). This is a chemical change.

Explanation:

Precipitation is when a chemical substance converts into a solid from a solution by converting the substance into an insoluble form or a super-saturated solution. When the reaction occurs in a liquid solution, the solid formed is called the precipitate.

Hope that helps.

3 0
3 years ago
The OBD-II catalytic converter monitor uses _______ sensor(s) to monitor catalytic converter efficiency.
oksano4ka [1.4K]
C comment for an explanation
8 0
3 years ago
At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 k?
photoshop1234 [79]
Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
\ln( \frac{K_2}{K_1} ) =  \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}  )
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2}  )=12.38
And so
\ln( \frac{K_2}{K_1})=12.38
And using K_1=6.1\cdot 10^{-8} s^{-1} , we find K2:
K_2=K_1 e^{12.38}=0.0145 s^{-1}


5 0
3 years ago
I dont know how to answer the last part about the results
Evgen [1.6K]

Answer:

I don't know I'm sorry. wish I could help

5 0
2 years ago
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