Situations during a soccer game in which the ball is determined in or out of play

A ring of diameter 7.70 cm is fixed in place and carries a charge of 5.00 mC uniformly spread over its circumference. (a) How mu

MatroZZZ [7]

Answer:

3.974 Joule

Explanation:

Diameter of ring = 7.7 cm

a = Distance from the center = d/2 = 3.85 cm = 0.0385 m

Q = Charge = 5 mC

q = Charge to move = 3.4 mC

k = Coulomb constant = 9×10⁹ Nm²/C²

Work done will be equal to Potential energy when mass is at center

$U=\frac{kQq}{a}\\\Rightarrow U=\frac{9\times 10^9\times 5\times 10^{-6}\times 3.4\times 10^{-6}}{0.0385}=3.974\ J$

∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule

4 0

3 years ago

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A california condor is sitting on the top branch of a redwood tree and has a GPE of 12,100 J. If the condor has a mass of 8kg, w

DanielleElmas [232]

Answer:

william h. seward secured the purchase of alaska from:

Explanation:

3 0

3 years ago

Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea

Alexxx [7]

Answer:

acceleration are

hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

Em₀ = U = m g h

Final point. To get off the ramp

Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

Em₀ = Em_f

mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

v = w r

w = v / r

we substitute

mg h = ½ v² (m + I / r²)

v² = 2 gh $\frac{m}{m+ \frac{I}{r^2} }$

v² = 2gh $\frac{1}{1 + \frac{I}{m r^2} }$

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

v² = v₀² + 2 a L

where L is the length of the plane

v² = 2 a L

a = v² / 2L

we substitute

a = $g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }$

let's use trigonometry

sin θ = h / L

we substitute

a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

I = m r²

we look for the acerleracion

a₁ = g sin θ $\frac{1}{1 + \frac{mr^2 }{m r^2 } }$ 1/1 + mr² / mr² =

a₁ = g sin θ ½

b) solid cylinder

I = ½ m r²

a₂ = g sin θ $\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }$ = g sin θ $\frac{1}{1+ \frac{1}{2} }$

a₂ = g sin θ ⅔

c) hollow sphere

I = 2/3 m r²

a₃ = g sin θ $\frac{1}{1 + \frac{2}{3} }$

a₃ = g sin θ $\frac{3}{5}$

d) solid sphere

I = 2/5 m r²

a₄ = g sin θ $\frac{1 }{1 + \frac{2}{5} }$

a₄ = g sin θ $\frac{5}{7}$

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ $\frac{105}{210}$