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PtichkaEL [24]
3 years ago
15

For the reactions of ketone body metabolism, _______.

Engineering
1 answer:
Artemon [7]3 years ago
7 0

Answer:

Is

Explanation:

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View the picture below and then correctly answer the questions using the following words: Temperate Zone, Tropical Zone, Polar Z
Nadusha1986 [10]

Can you provide the picture? Thanks !

8 0
2 years ago
Read 2 more answers
Compared to 15 mph on a dry road, about how much longer will it take for
Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph        ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10  m/s ²                         ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0
3 years ago
Read 2 more answers
What does this work for
Anastaziya [24]

Answer:

it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter

Explanat

8 0
2 years ago
In order to fill a tank of 1000 liter volume to a pressure of 10 atm at 298K, an 11.5Kg of the gas is required. How many moles o
lesya [120]

Answer:

The molecular weight will be "28.12 g/mol".

Explanation:

The given values are:

Pressure,

P = 10 atm

  = 10\times 101325 \ Pa

  = 1013250 \ Pa

Temperature,

T = 298 K

Mass,

m = 11.5 Kg

Volume,

V = 1000 r

   = 1 \ m^3

R = 8.3145 J/mol K

Now,

By using the ideal gas law, we get

⇒ PV=nRT

o,

⇒ n=\frac{PV}{RT}

By substituting the values, we get

       =\frac{1013250\times 1}{8.3145\times 298}

       =408.94 \ moles

As we know,

⇒ Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}

or,

⇒        MW=\frac{m}{n}

                   =\frac{11.5}{408.94}

                   =0.02812 \ Kg/mol

                   =28.12 \ g/mol

3 0
2 years ago
What are the three most common types of relearn procedures?
WITCHER [35]

Answer:

The three types of relearn procedures are auto relearn, stationary and OBD.

Explanation:

In TPMS system, after the direct service like adjustment of air pressure, tire rotation or replacement of sensors etc, is performed then maximum vehicle  often needs TPMS system relearn that needs to be performed.

For performing these relearn procedure, there are mainly three types:

  1. auto relearn
  2. stationary relearn
  3. OBD

After applying the relearn process, the TPMS system will again be in proper function.

8 0
3 years ago
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