Answer:
1.) 0 kgm/s
2) 6.3 kg
3) -0.0978 m/s
4)
5)
6)
An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.
2) What is the mass of the third piece?
3) What is the x-component of the velocity of the third piece?
4) What is the y-component of the velocity of the third piece?
5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?
6) Calculate the increase in kinetic energy of the pieces during the explosion
Explanation:
Since explosions and collisions follow the law of conservation of Momentum.
1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system
Since the body was initially at rest,
Magnitude of the initial momentum of the system = 0 kgm/s
Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.
2) mass of the third piece
Sum of all the masses = 16.2 kg
4.7 + 5.2 + c = 16.2
c = 6.3 kg
3) doing an x-component balance on momentum
(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0
-0.6163 + 6.3v = 0
v = -0.0978 m/s
Hope this Helps!!!
Using the theorem of kinetic energy
1/2mVf² - 1/2mVi²= WF + Wp, Wp=0
WF = F. AB, AB=5m and F= 40N, m=20kg
so the final kinetic is KEf= 1/2mVf² = WF =<span>F. AB= 40*5=200J
</span>
the final velocity is 1/2mVf² <span>=200, implies Vf= sqrt(20)=2sqrt(5)m/s</span>
Answer:
Explanation:
The kinetic energy of an object is given by 
where 
is the mass of the object and 
is the velocity of the object.
We can set up the following equations with the information given:
For part B, we have the same equation, but kinetic energy is now 
.
Therefore:
.
Answer:
2.11 %
Explanation:
In acidic medium iron is in Fe ⁺³ oxidation state .
Equivalent weight = 56 / 3
= 18.33 gm
acid used in titration
= 16.37 mL of .0233 M
= 16.37 x .0233 mL of M soln
= .38 mL of M soln
.38 mL of M soln reacts with .3298 gm of ore
1000 mL of M soln = (.3298 / .38) x 1000 gm of iron ore
867.89 gm
This must contain one gm equivalent of iron
or 18.33 gm
867.89 gm of ore contains 18.33 gm of iron
mass % of iron in the given ore
= (18.33 / 867.89) x 100
= 2.11 %
Answer:
2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂
2Na + MgCl₂ → 2NaCl + Mg
Explanation:
A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation
Among the given chemical reactions, we have;
2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂
In the above reaction;
The number of phosphorus, P, on either side of the equation = 2
The number of bromine atoms, Br, on either side of the equation = 6
The number of chlorine atoms, Cl, on either side of the equation = 6
Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced
The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg
In the above reaction, there are two sodium atoms, Na, one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced
