Answer:
40 mm
Explanation:
=Proper Length of rod = 2.05 m
Speed of rod = 6.01×10⁷ m/s
Length contraction
Final Length will be 2.01 m
So length that is contracted is 2.05-2.01 = 0.04 m
The length is contracted by 40 mm.
Answer:
Explanation:
Mass of station wagon M = 1200 kg , velocity V = 12 m/s
Mass of car m = 1800 kg , velocity v = 20 m/s
a ) Let centre of mass of car and station wagon be at a distance d from wagon
Taking moment of weight about it
1200 x d = 1800 x ( 40 - d )
3000 x d = 1800 x 40
d = 24 m
b ) Total momentum = MV +mv
= 1200 x 12 + 1800 x 20
= 50400 kg ms⁻¹
c ) Speed of centre of mass
v₀ = (MV + mv )/(m+M)
= 50400/3000
= 16.8 m/s
d) System Total momentum = velocity of cm x total mass
= 16.8 x 3000 = 50400 m/s .
It is equal to what was calculated in part b)
Answer:
Explanation: Flow rate is how much volume is going through in an one second there for
Flow rate = Velocity × Cross sectional Area
Here cros sectional Area is = π r²
= π × 20²
= 1256 m²
Q = V ×A
8×10³ = V × 1256
V = 6.4 m/s
1. <span> Fp = Mg*sin24 = 0.407Mg. </span>
<span>Fn = Mg*Cos24 = 0.914Mg. </span>
<span>Fk = u*0.914Mg. </span>
<span>Fp-Fk = M*a. </span>
<span>0.407Mg-u*0.914Mg = M*a.
</span>Divide by Mg:
<span>0.407-0.914u = a/g = 3.1/-9.81, </span>
<span>0.407-.914u = -0.316, uk = 0.791.
</span>
2. <span> M*g = 1200*9.8 = 11,760 N. = Wt. of car. = Normal force(Fn). </span>
<span>Fp = Mg*sin 0 = 0. </span>
<span>Fk = uk*Fn = 0.1323*11,760 = 1556 N. </span>
<span>Fp-Fk = M*a. </span>
<span>0-1556 = 1200a, a = -1.30 m/s^2. </span>
<span>V^2 = Vo^2 + 2a*d. </span>
<span>0 = (20.1)^2 - 19.6d
</span>