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shutvik [7]
3 years ago
8

explain a method applied to test a diesel engine at constant speed at list the method used to measure the friction power​

Physics
1 answer:
lana66690 [7]3 years ago
3 0

Automotive idustries  made a paradigm shift in selection of viscometrics of engine lubricant,from higher to lower viscosity grade,for improving fuel economy

Explanation:

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30 km/h is _________m/s?<br> A.)8.3<br> B.)5.6<br> C.)13.9<br> D.)11.1
Juli2301 [7.4K]

Answer:

Correct  Option :-  A

Explanation:

3 0
2 years ago
Plz help i really need someone help
lesantik [10]

Answer:

i can't see the picture, it is blocked off, can you write down your question?

Explanation:

6 0
3 years ago
On an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates
11Alexandr11 [23.1K]

The acceleration due to gravity (g) on this planet is 39.44 m/s²

<h3>What is solar system?</h3>

Solar system consists of all the planets and the most importantly the center of the solar system is Sun.

Given is an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds.

The time period of the pendulum is

T = 2π √l/g

Squaring both sides, we get

l/g = T² / 4π²

g = 4π²l/ T²

Substitute Time period T = 2s and  length l = 4m,  we get

g = 4π²x 4/ 2²

g =39.44 m/s²

Thus, the acceleration due to gravity on this planet is 39.44 m/s²

Learn more about solar system.

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3 0
2 years ago
(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
murzikaleks [220]

Answer:

a) t = -\frac{ln(2)}{k}

b) See the proof below

A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

\frac{1}{2} = e^{kt}

Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

For this case we want to find the value of t when we have remaining \frac{A_o}{8}

So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

\frac{1}{8} = 2^{-\frac{t}{T}}

We can apply natural log on both sides and we got:

ln(\frac{1}{8}) = -\frac{t}{T} ln(2)

And if we solve for t we got:

t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

8 0
3 years ago
Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons.
nikklg [1K]

Answer:

Sodium formate is the sodium salt of formic acid which is given as HCOONa.

Explanation:

The basic structure of Sodium formate consists of following bonds:

  1. The main Ionic bond between the HCOO^- radical and Na^+.
  2. The  sigma covalent bonds between atom of H, atom of C and  both atoms of O.
  3. The pi bond between atom of C and one atom of O.

The structure along with lone pairs is given as attached

7 0
3 years ago
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