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Digiron [165]
3 years ago
15

How do liquid water, ice, and water vapor differ from each other?

Physics
2 answers:
True [87]3 years ago
6 0
<span>They are made of different states of matter.</span>
anzhelika [568]3 years ago
5 0
Liquid water is a liquid Ice is a solid Water vapour is a gas
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There are ___________ of galaxies in the universe containing __________ of stars in each galaxy.
aliya0001 [1]
<span>There are Billions and billions of galaxies in the universe containing Trillions and trillions of stars in each galaxy.</span>
8 0
3 years ago
Read 2 more answers
Which of the following statements is FALSE?
kobusy [5.1K]
The false statement would be :
C. constructive interference occurs when the crest of one wave overlaps the trough of another and their individual effects are reduced or cancelled out.

Constructive interference should be when the crest of one wave overlaps another and their individual effects are add up, becoming a single wave of increased amplitude<span />
6 0
3 years ago
A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup
ohaa [14]
Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) =  7.5 mol

As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)

When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V= \frac{(7.5 \, mol)*(8.314 \, J/(mol-K))*(300 \, K)}{(92205.75 \, Pa)} =0.2029 \, m^{3}

V = (0.2029 m³)*(10³ L/m³) = 202.9 L

Answer: 203 liters
3 0
3 years ago
A ball is thrown so that its initial vertical and horizontal components of velocity are 30 m/s and 15 m/s, respectively. Estimat
mihalych1998 [28]

Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

6 0
2 years ago
At what speed would a 3.00 x 10^4 kg airplane have to fly and with a momentum of 1.60 x 10^9 kg.m/s
Ymorist [56]

Answer:

5.3×10⁴ m/s

Explanation:

From the question,

Momentum = mass× velocity

M = mV................ Equation 1

Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane

make V the subject of the equation

V = M/m.................. Equation 2

Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg

Substitute these values into equation 2

V = 1.6×10⁹/3.0×10⁴

V = 5.3×10⁴ m/s

3 0
3 years ago
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