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4 years ago

What factor determines how much a particular color refracts as it passes through a prism?

Physics

2 answers:

inysia [295]4 years ago

7 0

Wave length B is the answer.

frutty [35]4 years ago

6 0

The correct answer is the wavelength of the light. Good luck m8

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A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

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4 0

4 years ago

Read 2 more answers

In 2-3 complete sentences, explain the relationship between speed, frequency, and wavelength.

Ahat [919]

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength. ... The frequency is the same as that of the source and is the number of waves that pass a point per unit time.

7 0

3 years ago

A uniform electric field of magnitude 112 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

notsponge [240]

Answer:

$1.37\cdot 10^5 m/s$

Explanation:

The beam of positively charged particles would not be deflected if the electric force acting on them is equal to the magnetic force:

$F_E = F_B\\qE = qvB sin \theta$

where

q is the charge of the particles

E is the magnitude of the electric field

v is the speed of the particles

B is the intensity of the magnetic field

$\theta$ is the angle between the direction of v and B

Since the beam of particles travels perpendicular to the magnetic field, we have $\theta=90^{\circ}, sin \theta=1$ . So, we can rewrite the equation as

$v=\frac{E}{B}$

where

$E=112 kV/m = 1.12\cdot 10^5 V/m\\B=0.82 T$

Substituting into the formula,

$v=\frac{1.12\cdot 10^5 V/m}{0.82 T}=1.37\cdot 10^5 m/s$

4 0

3 years ago

A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center. The linear speed of a passenger on the

Marina86 [1]

Answer:

a) The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 90^{\circ}$ .

b) Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 270^{\circ}$ .

c) The Ferris wheel takes 14.646 seconds to make a revolution.

Explanation:

a) An object that rotates at constant angular velocity reports a centripetal acceleration and no tangential acceleration. When passenger passes through the lowest point in her circular motion, centripetal acceleration goes up to the center.

In addition, centripetal acceleration is determined by the following expression:

$a_{R} = \frac{v^{2}}{R}$ (Eq. 1)

Where:

$a_{R}$ - Centripetal acceleration, measured in meters per square second.

$v$ - Linear speed, measured in meters per second.

$R$ - Radius of the Ferris wheel, measured in meters.

If we know that $v = 6\,\frac{m}{s}$ and $R = 14\,m$ , the magnitude of radial acceleration is:

$a_{R} = \frac{\left(6\,\frac{m}{s} \right)^{2}}{14\,m}$

$a_{R} = 2.571\,\frac{m}{s^{2}}$

The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 90^{\circ}$ .

b) In the highest point the magnitude of radial acceleration is the same but direction is the opposed to that at lowest point. That is, centripetal acceleration goes down to the center.

Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$ , $\angle = 270^{\circ}$ .

c) At first we need to calculate the angular velocity of the Ferris wheel ( $\omega$ ), measured in radians per second, by using the following expression: