Answer:
<h3>
The volume needed to expand the gas is = 11 Lit = 11000 </h3>
Explanation:
Given :
Initial volume Lit.
Energy J = 9.38 Lit × atm. ⇒ ( 1 Lit×atm. = 101.325 J )
Pressure 650 torr = 0.855 atm. ⇒ ( 1 torr = 0.00132 atm )
In this example we have to be aware of unit conversion system.
From the laws of thermodynamics,
Here in this example, all the energy of combustion is converted into work to push back the piston
Lit =
According to Newton's second law of motion, acceleration of an object depends upon the mass of the object and the force acting on it
Explanation:
- The equation for the second law is F = Ma, where m is the mass of the object, a is the acceleration and F is the force acting on it
- Newton's second law of motion is applied in many situations for example, if you apply the same amount of force to pull a car and a truck. We all know that the car will move and the truck will not because the car has less mass than the truck. So, the truck needs more force to apply for it to move.
This is not as simple as it looks.
What quantity are we going to compare between the two cases ?
Yes, I know ... the "amount of work". But how to find that from the
numbers given in the question ?
Is it the same as the change in speed ?
Well ? Is it ?
NO. IT's NOT.
In order to reduce the car's speed, the brakes have to absorb
the KINETIC ENERGY, and THAT changes in proportion to
the SQUARE of the speed. ( KE = 1/2 m V² )
Case 'A' :
The car initially has (1/2 m) (100²)
= (1/2m) x 10,000 units of KE.
It slows down to (1/2 m) x (70²)
= (1/2m) x 4,900 units of KE.
The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.
Case 'B' :
The car initially has (1/2 m) (79²)
= (1/2m) x 6,241 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 6,241 units of KE.
Just as we suspected when we first read the problem,
the brakes do more work in Case-B, bringing the car
to a stop from 79, than they do when slowing the car
from 100 to 70 .
But when we first read the problem and formed that
snap impression, we did it for the wrong reason.
Here, I'll demonstrate:
Change Case-B. Make it "from 70 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (70²)
= (1/2m) x 4,900 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 4,900 units of KE.
-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.
-- Then, to slow the whole rest of the way from 70 to a stop,
the brakes absorbed only 4,900 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr
than to slow it the whole remaining 70 km/hr.
That's why you can't just say that the bigger change in speed
requires the greater amount of work.
______________________________________
It works exactly the same in the opposite direction, too.
It takes less energy from the engine to accelerate the car
from rest to 70 km/hr than it takes to accelerate it the
next 30, to 100 km/hr !
Answer & Explanation:
volume of the block = 4.5 x 5 x 6 = 135 cm^3 also equal to 1.35 x 10^-4 m^3
USE THE FORMULA
<em>Density = mass / volume</em>
<em><u /></em>
<em><u>Answer in gcm^-3</u></em>
<em>Density = 1502 / 135</em>
<u>Density = 11.1 gcm^-3</u>
<em><u /></em>
<em><u>Answer in kgm^-3</u></em>
<em>Density = 1.502 / 1.35 x 10^-4</em>
<u>Density = 11125.9 kgm^-3</u>
Elements to the left tend to form positive ions is the right answer mark me brainlist