1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Leviafan [203]
3 years ago
8

The freezing point of water is a (physical/chemical) property. The ability of oxygen to react with iron to cause rust is a (phys

ical/chemical) property.
Physics
2 answers:
AleksandrR [38]3 years ago
7 0
So the freezing point of water is physical and the ability of oxygen to react with iron to cause rust is chemical
Svetllana [295]3 years ago
6 0

The freezing Point of water is a physical property

The ability of oxygen to react with iron to cause rust is chemical property

Explanation:

You might be interested in
An object moves with constant acceleration 3.10 m/s2 and over a time interval reaches a final velocity of 12.4 m/s. If its initi
Harrizon [31]

Answer:

Explanation:

Given:

a = 3.10 m/s^2

vf = 12.4 m/s

vi = -6.2 m/s

t = (vf - vi)/a

= (12.4 + 6.2)/3.1

= 6 s

displacement = (vf - vi)*t

= (12.4 + 6.2) * 6

= 111.6 m.

3 0
3 years ago
Read 2 more answers
A person with a mass of 72 kg is riding a bicycle is accelerating at a rate of 5m/s on a horizontal surface. What is the weight
nikklg [1K]

Answer:

w = 706.32 [N]

Explanation:

The force due to gravitational acceleration can be calculated by means of the product of mass by gravitational acceleration.

w = m*g

where:

w = weight [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

Then we have:

w = 72*9.81\\w = 706.32 [N]

7 0
3 years ago
Can water and wind change the shape of a mountain
Art [367]

Answer:

Yes, through erosion.

Explanation:

Water, wind, and ice shape earths surface. Water, wind, & ice move sediment to another area this process is called erosion.

Mark me brainliest, hope this helps

4 0
3 years ago
There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center)
Aneli [31]

Answer:

q₁ =± 1.30 10⁻⁶ C  and   q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

    F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

    F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

    F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

    q₁ + q₂ = 2 q

Let's replace

    F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

    F1 = k q₁ q₂ / r²

    F2 = ½ k (q₁ + q₂) / r²    

    F1 r² / k = (q₁ q₂)

    F2 r² / k = (q₁ + q₂)/2

    q₁ = 2F2 r² / k - q₂

We substitute in the other equation

    F1 r² / k = (2F2 r² / k - q₂) q₂

    0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

    F1 r² / K = 0.0765 0.440² / 8.99 10⁹

    F1 r² / K = 1.65 10⁻¹²

   (2F2 r² / k) =2  0.10 0.44² / 8.99 10⁹

    (2F2 r2 / k) = 4.30 10⁻¹²

    q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

    q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

    q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

    q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

    F1 = k q₁ q₂ / r²

    q₁ = F1 r² / (k q₂)

    q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

    q₁ = 1.30 10⁻⁶ C

3 0
3 years ago
1. Earth is approximately a sphere of radius 6.37х106 m. what are (a) its circumference in kilometers, (b) its surface are in sq
Minchanka [31]

As we know that circumference of the sphere is given as

C = 4\pi R

here we know that

R = 6.37 \times 10^6 m

now we have

C = 4\pi (6.37 \times 10^6)

C = 8\times 10^7 m

C = 8 \times 10^4 km

PART B)

surface area of the sphere is given as

A = 4\pi R^2

R = 6.37 \times 10^3 km

A = 4\pi (6.37\times 10^3)^2

A = 5.1 \times 10^8 km^2

PART C)

Volume of the sphere is given as

V = \frac{4}{3}\pi R^3

here we have

V = \frac{4}{3}\pi(6.37 \times 10^3)^3

V = 1.1 \times 10^{12} km^3

5 0
4 years ago
Read 2 more answers
Other questions:
  • The equation P=20sin(2πt)+40 models blood pressure, P, where t represents time in seconds. Find the blood pressure after 21 seco
    5·1 answer
  • How does under water welding work?
    15·2 answers
  • A metal spoon can be classified as which of the following
    13·1 answer
  • The total number of stars in the observable universe is roughly equivalent to:
    10·1 answer
  • A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.
    5·1 answer
  • A sprinter must average 24.0 mi/h to win a 100-m dash in 9.30 s. What is his wavelength at this speed if his mass is 84.5 kg?
    8·2 answers
  • Which of the following would produce a star with the longest lifespan
    9·1 answer
  • Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially hel
    9·1 answer
  • Giúp với mng ơi yêu mng <3 béo ị
    5·1 answer
  • A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!