Answer:
1400 N
Explanation:
Change in momentum equals impulse which is a product of force and time
Change in momentum is given by m(v-u)
Equating this to impulse formula then
m(v-u)=Ft
Making F the subject of the formula then

Take upward direction as positive then downwards is negative
Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then

Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = 
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
<u>Determine the molar conductivity of acetic acid</u>
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
Answer:
The correct option is A
Explanation:
From the question we are told that
The mass number is 
Generally the mean radius is mathematically evaluated as

Here
is a constant with a value 
So



Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
Answer:
yes it doesn't matter
Explanation:
it doesn't matter because troughs and crests are the same and either can be used