Answer:
-1.43 m/s relative to the shore
Explanation:
Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:
where 
are the mass of the swimmer and raft, respectively. 
are the velocities of the swimmer and the raft after the run, respectively. We can solve for 
So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore
Step 2: Use the slope to find<span> the y-intercept. </span>Line<span> is </span>parallel<span> so use m = 2/5. </span>6<span>. </span>Find<span>the </span>equation<span> of a </span>line passing through the point<span> (8, –</span>9<span>) perpendicular to the </span>line<span> 3x + 8y = 4.</span>
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
The density of the flow at the exit is 2.2721 kg/m³
In exothermic reactions, heat and light are released to the surrounding environment. On the other hand, in an endothermic reaction, heat is required and therefore it can be considered as a reactant.
- In exothermic reactions, light and heat are released into the environment (Option D).
- Exothermic reactions release energy in the form of heat or light.
- Combustion reactions are generally exothermic reactions.
- After an exothermic reaction takes place it is possible to observe that the energy of the products of the reaction is lesser than the energy of the reactants.
- The energy released in exothermic reactions is evidenced by the increase in temperature of the reaction.
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Answer:
Explanation:
Initial kinetic energy of the system = 1/2 mA v0²
If Vf be the final velocity of both the carts
applying conservation of momentum
final velocity
Vf = mAvo / ( mA +mB)
kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²
= mA²vo² / 2( mA +mB)
Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6
mA v0² x ( mA +mB) / mA²vo² = 6
( mA +mB) / mA = 6
mA + mB = 6 mA
5 mA = mB
mB / mA = 5 .
