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nikklg [1K]
3 years ago
11

A 2.30-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. a solid sph

ere of mass 0.25 kg is thrown horizontally with a speed v1 = 12.5 m/s to hit the rod at the point a one-fifth of the way up from the bottom of the rod. the sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle θ before swinging back toward its original position. what is the angular speed of the rod immediately after the collision?
Physics
1 answer:
Marina86 [1]3 years ago
8 0

Solution:


initial sphere mvr = final sphere mvr + Iω 
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m² 
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω 
where: ω = 2.87 rad/s 

So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)² 
E = 12.64 J becomes PE = mgh, so 
12.64 J = 2.3 kg * 9.8m/s² * h 
h = 0.29 m 

h = L(1 - cosΘ) → where here L is the distance to the CM 
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ 
Θ = arccos((1-0.29)/1) = 44.77 º 

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Answer:

Explanation:

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5 0
3 years ago
Explain how the potential energy of two charged particles depends on the distance between the charged particles and on the magni
maks197457 [2]

Answer:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges

Explanation:

According to Coulomb's Law, the potential energy of two charged particles is directly proportional to the product of the two charges and inversely proportional to the distance between the charges.  Since the potential energy  of two charged particles is directly proportional to the product of the two charges, its magnitude increases as the charges of the particles increases. For like charges, the potential energy is positive(the product of the two alike charges must be positive) and since potential energy is inversely proportional to the distance between the charges therefore it decreases as the particles get farther apart . For opposite charges, the potential energy is negative(the product of the two opposite charges must be negative) and since potential energy is inversely proportional to the distance between the two charges, it becomes more negative as the particles get closer together.

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3 years ago
A static charge that is produced by a nearby objet is an?
Wittaler [7]
I believe that it is electric field
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3 years ago
The low-frequency speaker of a stereo set has a surface area of 0.06 m2 and produces 2.03 W of acoustical power. What is the int
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Answer:

33.83W/m²

Explanation:

The intensity of the speake at the surface is

I = P/A

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3 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
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