an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respectively. if the energy input is 3600 joule. what is the work done by the engine

Graphite is great lubricant because each carbon atom forms weak covalent bonds with three other carbon atoms.

Diamond and graphite both are entirely made up of carbon but both the elements are completely different because of the bonding of carbon atoms. In graphite the carbon atoms form weak covalent with the other three carbon atom due to which there is less or no force between the atoms, and makes the graphite very soft and slippery, and graphite act as lubricant.

Answer:

Carbon atoms form strong bonds within each graphite layer but weak bonds between layers.

Explanation:

Graphite is known as solid lubricant which means it is a lubricant which is in the solid state form.

Here we know that lubricant has function to make the relative movement very smooth or easy.

So here in structure of graphite it is formed in such a way that all carbon atoms in graphite layer are bonded strongly with each other. While two layers of graphite are weakly bonded to each other

So here two layers of graphite can easily slide over each other which is useful for the function of lubricants

So correct answer will be

Carbon atoms form strong bonds within each graphite layer but weak bonds between layers.

T = half life of cobalt-60 = 5.3 years

λ = decay constant for cobalt-60 = ?

decay constant is given as

λ = 0.693/T

inserting the values

λ = 0.693/5.3 = 0.131

N₀ = initial amount of cobalt-60 = 10 g

t = time of decay = 21.2 years

N = final amount of cobalt-60 after time "t"

final amount of cobalt-60 after time "t" is given as

N = N₀ [tex]e^{-\lambda t}[/tex]

inserting the values

N = 10 [tex]e^{-(0.131) (21.2)}[/tex]

N = 10 [tex]e^{-(0.131) (21.2)}[/tex]

N = 10 x 0.0622 g

N = 0.63 g

hence

C. 0.63 g

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²

Final answer:

The magnitude of the car's acceleration is calculated using the change in velocity over the time taken. With a change in velocity of 12.0 m/s over 8.0 seconds, the car's acceleration is 1.5 m/s².

Explanation:

To find the magnitude of the car's acceleration, we can use the formula for acceleration, which is the change in velocity (Δv) divided by the time (Δt) it takes for that change. In this case, the car's velocity increases from 18.0 m/s to 30.0 m/s over 8.0 seconds.

We calculate acceleration (a) as follows:

a = Δv / Δt

Δv = final velocity - initial velocity = 30.0 m/s - 18.0 m/s = 12.0 m/s

Δt = 8.0 s

Thus, a = 12.0 m/s / 8.0 s = 1.5 m/s².

The magnitude of the car's acceleration is 1.5 m/s².

POWER OF CIRCUIT

P=VI

While,

VOLTAGE OF CIRCUIT

V=IR

THEN,

P=I^R

P/R=I^

2.00/30.0=I^

I=0.258A.

Answer:0.258

Explanation:

Given data

          mass (m) = 0.2 Kg ,

       velocity (v) =300 m/s ,

      calculate momentum (p) = m × v  

                                                = 0.2 × 300

                                              P  = 60 Kg. m/s

Momentum of the bullet is 60 Kg. m/s..