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2 years ago

What is empirical evidence and what's is it not

1 answer:

3 0

Empirical evidence is information acquired by observation or experimentation. This data is recorded and analyzed by scientists and is a central process as part of the scientific method.
As example of not empirical evidence:
Language, letters, words, truths, numbers, logic, mathematics, on and on.
All of these things do not have empirical existence. They exist only in the mind.

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If you were to mix 10 grams of salt into 100 grams of water, how much will the total mass of the solution be?

Ede4ka [16]

110 grams

mass of salt + mass of water

10 + 100

110g

4 0

1 year ago

An incident ray of light strikes a diamond at an angle of

lys-0071 [83]

Answer:

The angle of refraction is option b: 17°.

Explanation:

We can find the angle of refraction by using Snell's law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})$

Where:

n₁: is the index of refraction of the medium 1 (air) = 1.0003

n₂: is the index of refraction of the medium 2 (diamond) = 2.42

θ₁: is the angle of incidence = 45°

θ₂: is the angle of refraction =?

Hence, the angle of refraction is:

$sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{1.0003*sin(45)}{2.42} = 0.2922$

$\theta_{2} = 17 ^{\circ}$

Therefore, the correct option is b: 17°.

I hope it helps you!

4 0

3 years ago

Read 2 more answers

Which best describes why liquids needs a container when a solid does not

Vladimir [108]

Answer:because liquids does not stick togather so it spreads out and solids don't

Explanation:

4 0

3 years ago

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A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

2 years ago

A ball is dropped from a height of 20m. If its velocity increases uniformly at the rate 10m/s2 with what velocity and after what

belka [17]

Answer: 2 seconds

Explanation:

6 0

3 years ago

Read 2 more answers