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PolarNik [594]
2 years ago
13

My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

es having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) a. m 70 kg; h = 8 m; and d = 40 m b. m 70 kg; h 8 m; and d 20 m c. m 35 kg; h 4 m; and d 40 m d. m 7 kg; h 16 m; and d 20 m e. m-7 kg;h-4 m; and d=20 m f. m105 kg; h-6.0 m; and d 30.0 m
Physics
1 answer:
Tresset [83]2 years ago
8 0

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

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8 0
3 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
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Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

J=\frac{E}{\rho}      (1)

E: electric field E(t)=0.0004t2−0.0001t+0.0004

ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

J=\frac{I}{A}        (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}

Next, you replace for all variables:

I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A

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4 0
3 years ago
A block of mass m is compressed against a spring (spring constant kk) on a horizontal frictionless surface. The block is then re
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Answer:

A) True, B) False, C) False  and  D) false

Explanation:

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Let's look for mechanical energy

Initial

     Emo = Ke = ½ k Dx2

Final

     Em1= ½ m v12

     Emo = Em1

     ½ k Δx2 = ½ m v₁²

    v₁² = k / m Δx²

    v₁ = √ k/m   Δx

Now let's calculate the speed when it falls

   Vfy² = Voy² - 2gy

   Vfy² = - 2gy

   Vf² = v₁² + vfy²

A) True     v₁ = A Δx

.B) False. As there is no rubbing the mechanical energy conserves

.C) False the velocity is proportional to the square root of the height

     v2y = v2 √2

. D) false promotional compression speed

3 0
3 years ago
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