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3 years ago

14

How is temperature and pressure related to air volume?

1 answer:

Solnce55 [7]3 years ago

8 0

The colder it is, the more calm the air particles are. They will often tend to be packed next to each other, therefore containing more pressure. However, the hotter it is, the faster and more actively the particles will move around, therefore, having less pressure.

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PHYSICS, WORTH 10 POINTS!

11Alexandr11 [23.1K]

Answer:

Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle. Joining the points together reveals the path of the magnetic field lines.

Explanation:

5 0

2 years ago

A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the object's density in units of g/ml?

Zinaida [17]

G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
2.20 cm, 1.35 cm, and 1.25 cm
Then the total volume is: V = lwh = 3.7125 cm^3
To get the density, we divide mass by volume: 2.50 g / 3.7125 cm^3 = 0.6734 g/cm^3 = 0.6734 g/mL

4 0

3 years ago

1. Faça as transformações:

zhannawk [14.2K]

Answer:

$seconds (s) = hours (h) *3,600 ; h = \frac{s}{3,600} \\g = kg * 1,000; kg = \frac{g}{1,000} \\cm = \frac{m}{100};m = cm * 100$

1. a) 0.5 h = 1,800 s

h) 20 cm = 0.2 m

b) 2.0 h = 7,200 s

i) 5.0 kg = 5,000 g

c) 3.5 h = 12,600 s

j) 1.5 kg = 1,500 g

d) 1/4 h = 900 s

k) 450.0 g = 0.45 kg

e) 3.0 m = 300 cm

l) 20.0 g = 0.02 kg

f) 2.5 m = 250 cm

m) 500.0 g = 0.5 kg

g) 0.5 m = 500 mm

n) 1000.0 g = 1 kg

3 0

2 years ago

A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th

s344n2d4d5 [400]

Answer:

15 cm

Explanation:

$h_{o}$ = Diameter of the coin = 15 mm

$h_{i}$ = Diameter of the image of coin = 5 mm

$d_{o}$ = distance of the coin from mirror = 15 cm

$d_{i}$ = distance of the image of coin from mirror = ?

Using the equation

$\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}$

$\frac{d_{i}}{15} = \frac{- (5)}{15}$

$d_{i}$ = - 5 cm

$R$ = radius of curvature

Using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}$

$\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}$

$R$ = - 15 cm

6 0

2 years ago

Light is incident on an air-water interface at an angle of 30 degree to the normal. What angle does the refracted ray make with

Vikentia [17]

Answer:

22 degree

Explanation:

Angle of incidence, i = 30 degree

the refractive index of water with respect to air is 4/3.

As the ray of light travels from rarer medium to denser medium, that mean air to water, the refraction takes place.

According to Snell's law,

Refractive index of water with respect to air = Sin i / Sin r

Where, r be the angle of refraction

4 / 3 = Sin 30 / Sin r

0.75 = 2 Sin r

Sin r = 0.375

r = 22 degree

Thus, the angle of refraction is 22 degree.

6 0

2 years ago