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charle [14.2K]
3 years ago
15

Visible light is a range of _____ energy EM waves in the electromagnetic spectrum that the human eye can see. A. zero B. high C.

low D. medium
Physics
2 answers:
patriot [66]3 years ago
7 0
The answer is a or d

vfiekz [6]3 years ago
4 0

The visible light falls into the measure of low energy range in the electromagnetic spectrum.  

Answer: Option C

<u>Explanation:</u>

The visible light is one of the 7 kind of electromagnetic spectrum that we know. Of those 7 kind, the visible light falls in between of the electromagnetic spectrum according to the wavelength of the light.

The visible light consists of VIBGYOR where the red has the 1.8 eV of the energy with violet to have around 3.4 eV. Therefore the correct answer is low.

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A bullet of mass 0.1 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 3 kg that is sitting at re
Xelga [282]

Answer:

(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s

(b) the kinetic energy of the bullet plus the block before the collision is 500J

(c) the kinetic energy of the bullet plus the block after the collision is 16.13J

Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

Part (A)

Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the block after the bullet embeds itself in the block

(0.1 x 100) + (3 x 0) = v (0.1 + 3)

10 = 3.1v

v = 10/3.1

v = 3.226 m/s

Part (B)

Initial Kinetic energy

Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki =  ¹/₂(0.1 x 100²) +  ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

Part (C)

Final kinetic energy

Kf = ¹/₂m₁v² + ¹/₂m₂v²

Kf = ¹/₂v²(m₁ + m₂)

Kf = ¹/₂ x 3.226²(0.1 + 3)

Kf = ¹/₂ x 3.226²(3.1)

Kf = 16.13 J

6 0
3 years ago
A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl
zysi [14]

Answer:\alpha =9.302\ rad/s^2

Explanation:

Given

mass of sphere m=250\ gm

diameter of sphere d=4.30\ cm

radius r=\frac{4.30}{2}\ cm

f=0.0200\ N

friction will provide resisting torque so

f\times r=I\times \alpha

where I=\text{moment of Inertia}

f=\text{friction force}

\alpha =\text{angular acceleration}

I=\frac{2}{5}mr^2

0.02\times r=\frac{2}{5}mr^2\times \alpha

\alpha =\frac{5}{2r}\times f

\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02

\alpha =9.302\ rad/s^2

(b)time taken to decrease its rotational speed by 21\ rad/s

t=\dfrac{\Delta \omega }{\alpha }

t=\dfrac{21}{9.302}

t=2.25\ s

6 0
3 years ago
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