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3 years ago

6

Why the speed of light decreases as it passes from air into another substance?

1 answer:

vladimir1956 [14]3 years ago

5 0

Answer:

If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.

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Which are dangerous places to put a flammable substance? Check all that apply.

Murrr4er [49]

Answer:

near a flame and a hot plate

Explanation:

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3 years ago

Fluid clutches are used on equipment that is subjected to?

mamaluj [8]

Fluid clutches are used on equipment that is subjected to higher torque per unit volume. This type of clutch makes use of an incompressible fluid like oil in order to transfer the input of a movement from a pedal and cylinder to an actuating cylinder. Releasing the this clutch would stop the transfer of power but still allowing the engine to turn continuously.

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3 years ago

A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome

Leya [2.2K]

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

$L=I\omega$

$L=\dfrac{1}{2}mr^2\omega$

Put the value into the formula

$L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320$

$L=12.96\ kg m^2/s$

(B). Rotation momentum of sphere is same rotational momentum of the flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

$L_{sphere}=L_{flywheel}$

$I\omega_{sphere}=I\omega_{flywheel}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}$

$\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}$

Put the value into the formula

$\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}$

$\omega_{sphere}=400\ rad/s$

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

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3 years ago

A 0.03 in.diameter glass tube is inserted into SAE 30 oilat 60°F. The contact angle (i.e., the angle a liquid forms in contact w

Snowcat [4.5K]

Answer:

The value of capillary rise in the tube = 0.0363 cm

Explanation:

Diameter of the glass tube = 0.03 meter

Contact angle = 22 degree

We know that the capillary rise (h) in the tube is given by the formula =

h = (2 α cos β) / (d g R )

⇒ d = density of gasoline = 749 kg / $m^{3}$

β = 22 degree

R = radius of the tube = 0.015 m

α = surface tension of gasoline = 0.0216 N / m

cos 22 = 0.927

Put all the values in the above formula we get

⇒ h = (2 × 0.0216 × 0.927) / (749 × 9.81 × 0.015)

⇒ h = 0.0363 cm

This is the value of capillary rise in the tube.

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3 years ago

What is the position at 2 seconds?

Step2247 [10]

5

If you go over 2 on the time bar, and you look for the black dot above the 2 on the bottom time bar, it would go up to 5.

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4 years ago