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blondinia [14]
2 years ago
9

The rotating blade of a blender turns with constant angular acceleration 1.60 rad/s2.

Physics
1 answer:
amm18122 years ago
7 0

Answer:

a) 23.1 s

b) 68.1 rev

Explanation:

a) t = ω/α = 37.0 / 1.60 = 23.125 ≈ 23.1 s

ω₁² = ω₀² + 2αθ

θ = (ω₁² - ω₀²) / 2α = (37.0² - 0.00²) / 2(1.60) = 427.8125 radians

427.8125 rad / 2π rad/rev = 68.08847...

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A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
2 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

5 0
3 years ago
How many grams of CO2 are in 10 mol of the compound?
Zepler [3.9K]

For our problem, this means that one mole of CO2 has a mass of 44.01 grams. So 22 grams divided by 44.01 grams is roughly 0.5 moles of CO2.


hope it helps

4 0
3 years ago
The spring of a toy car is wound by pushing the car
Zepler [3.9K]

The elastic potential energy stored in the car's spring during the process is 3.75 J

<h3>Determination of the spring constant</h3>

From the question given above, the following data were obtained:

  • Force (F) = 15 N
  • Extention (e) = 0.5 m
  • Spring constant (K) =?

K = F/e

K = 15 / 0.5

K = 30 N/m

<h3>Determination of the potential energy</h3>
  • Spring constant (K) = 30 N/m
  • Extention (e) = 0.5 m
  • Potential energy (PE) =?

PE = ½Ke²

PE = ½ × 30 × 0.5²

PE = 15 × 0.25

PE = 3.75 J

Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J

Learn more about energy stored in spring:

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7 0
2 years ago
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How do you disseminate a car if you rub it with rubber
natta225 [31]
I have honestly never read anything about a car being disseminated,
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5 0
3 years ago
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