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3 years ago

If you want to study how energetic waves affect matter, you should study waves with a _______.

1 answer:

4 0

Higher Frequency and shorter wavelength. The energy of light waves increases when there is an increasing frequency and a higher frequency means there are shorter wavelengths. The equation lambda=c/f where lambda is wavelength and f is frequency and c is the speed of the wave.

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Identify the correct order of increasing intermolecular force of attraction for the three most common states of matter.

professor190 [17]

I believe it’s C) gas- liquid- solid

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4 years ago

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Why would you expect sodium (Na) to react strongly with chlorine (Cl)? 1. They both need to lose one electron.2. They both need

vladimir1956 [14]

Give me some answer choices and i will be happy to help

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3 years ago

A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular

AURORKA [14]

Answer:

A) ω₂ = 28 rev/min

B) ω_3 = 28 rev/min

Explanation:

A) Initial moment of inertia (I₁) is the rod's own ((1/12)ML²) plus that of the two rings (two lots of mr² if we treat each ring as a point mass).

Thus;

I₁ = ((1/12)ML²) + 2(mr²)

Where;

M is mass of rod = 0.03 kg

L is length of rod = 0.4m

m is mass of small rings = 0.02 kg

r is radius of small rings = 0.05 m

Thus;

I₁ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.05²)

I₁ = 0.0021 kg.m²

Now, when the rings slide and reach the ends of the rod i.e at r = L/2 = 0.4/2 = 0.2m from axis, the new Moment of inertia is;

I₂ = ((1/12) x 0.03 x 0.4²) + 2(0.02 x 0.2²)

I₂ = 0.0036 kg.m²

From conservation of angular momentum, we know that:

I₁ω₁ = I₂ω₂

We are given ω₁ = 48 rev/min.

Thus; plugging in the relevant values;

0.0021 x 48 = 0.0036ω₂

0.1008 = 0.0036ω₂

ω₂ = 0.1008/0.0036

ω₂ = 28 rev/min

b) For this, we have the same scenario as the case above where the ring just reaches the ends of the rods.

Thus,

I₂ω₂ = I_3•ω_3

So,0.0021 x 48 = 0.0036ω_3

ω_3 = 28 rev/min

7 0

3 years ago

What is the average velocity in the time interval 3 to 4 seconds

AlekseyPX

Since the position doesn't change over that time, it's zero

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4 years ago

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A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2 /3. Suppose a 2.00-m r

Stels [109]

To solve this problem we will apply the concepts related to the moment of inertia and Torque, the latter both its translational and rotational expression.

According to the information given the moment of inertia of the body would be

$I = \frac{1}{3} mL^2$