Answer:
c) L³/T³
Explanation:
If t stands for time, the units are:
(V) = L³, (t) = T
The units for the equation:
V(t) = At³
must be:
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Answer:
3141N or 3.1 ×10³N to 2 significant figures. The can experiences this inward force on its outer surface.
Explanation:
The atmospheric pressure acts on the outer surface of the can. In order to calculate this inward force we need to know the total surface area of the can available to the air outside the can. Since the can is a cylinder with a total surface area given by 2πrh + 2πr² =
A = 2πr(r + h)
Where h = height of the can = 12cm
r = radius of the can = 6.5cm/2 = 3.25cm
r = diameter /2
A = 2π×3.25 ×(3.25 + 12) = 311.4cm² = 311.4 ×10-⁴ = 0.031m²
Atmospheric pressure, P = 101325Pa = 101325 N/m²
F = P × A
F = 101325 ×0.031.
F = 3141N. Or 3.1 ×10³ N.
Answer:
A) 199.78 J
B) 9.292x10^14 J
C) 4.2x10^7 m/s
D) 0.65 m
E) 1.13x10^-8 sec
D) 2.94x10^-9 sec
Explanation:
mass of ball = 0.0580 kg
A)
If smashed at v = 83.0 m/s, KE is
KE = 0.5mv^2
= 0.5 x 0.0580 x 83.0^2
= 199.78 J
B) if returned at v = 1.79×10^8 m/s, KE will be
KE = 0.5mv^2
= 0.5 x 0.0580 x (1.79×10^8)^2
= 9.292x10^14 J
C) during Einstein's return, velocity of rabbit relative to players is
Vr = 2.21×108 m/s
Rabbit's velocity relative to ball = 2.21×10^8 - 1.79×10^8
= 4.2x10^7 m/s
D) the rabbit's speed approaches the speed of light so we consider relativistic effect. The rabbit's measured distance is
l = l°( 1 - v^2/c^2)
= 2.5(1 - 2.21/3)
= 2.5 x 0.26
= 0.65 m
E) according to the players, the time taken by the rabbit is
t = d/v = 2.5/ 2.21×10^8
= 1.13x10^-8 sec
F) the time for rabbit as measured by rabbit is relativistic
t = t°( 1 - v^2/c^2)
= 1.13x10^-8 (1 - 2.21/3)
= 1.13x10^-8 x 0.26
= 2.94x10^-9 sec
