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labwork [276]
3 years ago
12

If V is a velocity, l a length, and v a fluid property having dimensions of L2T-1, which of the following combinations are quali

tatively dimensionless (show work for a through d):(a) V l v(b) Vl/v(c) V2v(d) V / (l v)
Physics
1 answer:
olasank [31]3 years ago
4 0

Answer:

(b) Vl/v

Explanation:

The dimension for V is LT-1

The dimension for Length is L

The dimension for time is T

The fluid property v has a dimension of L2T-1, which is equal to square of the length divided by time.

for (a) Vlv

LT-1×L× L2T-1

= L^4T^-2

(b) Vl/v

LT^-1 ×L/ L^2 T^-1

=1. ; this is dimensionless.

(c)V2v

(LT^-1)^2 × L^2 T^-1

= L^4 T^-3

(d) V/(lv)

LT^-1/(LL^2 T^-1)

= L^-2

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when the momentum of the vehicle moving at 30 km/h is higher than the one from the vehicle moving at 60 km/h

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2 years ago
After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?
faust18 [17]

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Explanation:

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8 0
4 years ago
A uniform disk has a moment of inertia that is (1/2)MR2. A uniform disk of mass 13 kg, thickness 0.3 m, and radius 0.2 m is loca
kicyunya [14]
The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
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6 0
4 years ago
Read 2 more answers
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
3 years ago
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