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AleksandrR [38]
2 years ago
10

What is the density of the baseball?

Physics
2 answers:
Georgia [21]2 years ago
7 0

Answer:

Chile the answer is yes

Explanation:

I just took the quiz

Aleksandr [31]2 years ago
5 0

The density of the baseball is 0.875 g/cm^3

Explanation:

The density of an object is given by the equation

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

For the baseball in this problem, we have:

m = 350 g is its mass

V=400 mL = 400 cm^3 is the volume

Substituting into the equation, we find the density:

d=\frac{350}{400}=0.875 g/cm^3

Therefore, the ball's density is outside the acceptable range.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

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A bus accelerates from 5.75 m/s at a rate of 1.25 m/s/s for 3.50
marysya [2.9K]

Answer:

10.125 meters?

Explanation:

Im taking 5.75m/s + 1.25 m/s/s (3.5) = my answer.

In those 3.5 seconds it travels 4.375.

I added that to 5.75 to get 10.125m

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3 years ago
Consider an experiment in which slow neutrons of momentum ¯hk are scattered by a diatomic molecule; suppose that the molecule is
madreJ [45]

Answer:

Check the explanation

Explanation:

When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.

position = amplitude x sine function(angular frequency x time + phase difference)

x = A sin(ωt + ϕ)

x = displacement (m)

A = amplitude (m)

ω = angular frequency (radians/s)

t = time (s)

ϕ = phase shift (radians)

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3 years ago
Consider 3 polarizers. Polarizer 1 has a vertical transmission axis and polarizer 3 has a horizontal transmission axis. Taken to
Nikolay [14]

Answer:

Detailed solution is given below:

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3 years ago
Find the final velocity of a 40 kg skateboarder traveling at an initial velocity of 10 m/s that moves up a hill from a height of
slavikrds [6]

Answer:

vf = 0

Explanation:

Since the initial height hi = 0, we can rewrite the energy equation as

vf^2 = vi^2 - 2ghf = (10 m/s)^2 - 2(10 m/s^2)(5 m) = 0

Therefore, his final velocity vf is

vf = 0

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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

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3 years ago
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