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aleksklad [387]
2 years ago
6

Brainliest if correct Question 1 of 10

Physics
1 answer:
Art [367]2 years ago
4 0

Answer:

B

Explanation:

hope this helps you

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This is the phase of matter with no fixed shape but fixed volume.
Nataly_w [17]
A liquid is a matter that neither has a fixed shape but fixed volume...in college really need to know this.

GOOD LUCK
6 0
3 years ago
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Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng
klemol [59]

Answer:

Speed:

2.01x10^{8}m/s

Wavelength:

4.24x10^{-7}m

Frequency:

4.74x10^{14}Hz

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

n = \frac{c}{v} (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

v = \frac{c}{n}

v = \frac{3x10^{8}m/s}{1.490}

v = 2.01x10^{8}m/s

Hence, the speed of the laser has a value of 2.01x10^{8}m/s

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.  

To determine the frequency it can be used the following equation

c = \nu \cdot \lambda  (2)

Where c is the speed of light, \nu is the frequency and \lambda is the wavelength

Then, \nu wil be isolated from equation 2.

\nu = \frac{c}{\lambda}  (3)

Before using equation 3 it is necessary to express \lamba in units of meters.

\lambda = 632.8nm . \frac{1m}{1x10^{9}nm} ⇒ 6.328x10^{-7}m

\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}

\nu = 4.74x10^{14}s^{-1}

\nu = 4.74x10^{14}Hz

Hence, the frequency of the laser has a value of 4.74x10^{14}Hz

Wavelength:

To determine the wavelength it can be used:

v = \nu \cdot \lambda

\lambda = \frac{v}{\nu}

Where v is the speed of the laser through the polystyrene.

\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}

\lambda = 4.24x10^{-7}m

Hence, the wavelength of the laser has a value of 4.24x10^{-7}m

3 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
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Plz help!
Strike441 [17]
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= 20m/s
Velocity= 20m/s
In 4seconds,distance covered=20x4= 80m
Therefore,an accident occurs because in 4seconds, the car would have moved past 20m
6 0
3 years ago
A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee
Olegator [25]
Quite low

Gravitation force = \frac{6.67* 10^{-11}*0.145*6.8 }{ 0.5^{2} } = 26.3 * 10^{-11} = 2.63 *10^{-10} N
6 0
3 years ago
Read 2 more answers
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