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Marina86 [1]
3 years ago
10

ANSWER FAST PLEASE. The bottom of a box has a surface area of 25.0 cm 2 . The mass of the box is 34.0 kilograms. Acceleration du

e to gravity at sea level is 9.80 m/s 2 . What pressure is exerted by the box where it rests?
A. 1.36 N/cm 2
B. 13.3 N/cm 2
C. 0.139 N/cm 2
D. 7.21 N/cm 2
Physics
2 answers:
Natasha_Volkova [10]3 years ago
7 0
34 * 9.8 = 333.2 
333.2 / 25 = 13.33

<u>B.13.3N/cm^2</u>
ohaa [14]3 years ago
5 0
13.3 is the correct answer!
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In the united states a standard letter sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for
Andre45 [30]

This question is incomplete

Complete Question

In the United States, a standard letter-sized piece of paper is 8.5 inches wide by 11 inches long. The international standard for a letter-sized piece of paper is different. The international standard is based on SI units: 21.0 cm wide by 29.7 cm long.

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

Answer:

a) 8.267721 inches ≈ 8.3 inches

b) 11.6929197 inches ≈ 11.7 inches

c) It's dimensions in inches for the international standard for letter - sized for paper = 8.3 wide inches by 11.7 inches long

d) The International standard letter - sized paper is longer.

Explanation:

a. Convert 21.0 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

21 cm =

Cross Multiply

21 cm × 0.393701inch/ 1 cm

= 8.267721 inches

Approximately 8.3 inches

b. Convert 29.7 cm to inches. Show your dimensional analysis setup.

1 cm = 0.393701inch

29.7 cm =

Cross Multiply

29.7 × 0.393701 inch/ 1 cm

= 11.6929197 inches

Approximately 11.7 inches

c. State the dimensions, in inches, of the international standard for a letter-sized piece of paper.

The international standard for a letter-sized has dimensions 21.0 cm wide by 29.7 cm long.

Where

21.0cm = 8.267721 inches

≈ 8.3 inches

29.7cm = 11.6929197 inches

≈ 11.7 inches

Hence, it's dimensions in inches = 8.3 inches by 11.7 inches.

d. Which piece of paper is longer: a U.S. letter-sized piece of paper, or an international letter-sized piece of paper?

U.S letter - sized paper = 8.5 inches wide by 11 inches long

International standard letter- sized paper = 8.3 wide inches by 11.7 inches long.

Hence, the International standard letter - sized paper is longer.

5 0
3 years ago
the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for
svetoff [14.1K]

One of the brightest nebulae in the night sky, the Orion Nebula may be seen with the unaided eye. The Trapezium is a young open cluster of four main stars in this magnitude 4 interstellar cloud of ionized atomic hydrogen.

<h3>What is the source of the Orion Nebula's crimson glow?</h3>
  • The hydrogen gas in the Orion Nebula, which is powered by radiation from young stars, gives off a crimson tint. The nebula's blue-violet regions are reflecting radiation from bright, blue-white O-type stars while the red areas are emitting light.
  • The Orion Nebula is one of many massive clouds of gas and dust in our Milky Way galaxy, say contemporary astronomers, and is one of the largest. It is approximately 1,300 light years away from Earth. This enormous hazy cocoon, which measures approximately 30 to 40 light-years in diameter, is generating potentially a thousand stars.  

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3 0
1 year ago
The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o
blsea [12.9K]

Answer:

F_f=840N

Explanation:

From the question we are told that

Weight of fireman W_f= 87.5kg

Pole distance D=4.10m

Final speed is V_f 1.75m/s

Generally the equation for velocity is mathematically represented as

v^2 = v_0^2 + 2 a d

Therefore Acceleration a

a'= v^2 / 2 d

a'= 0.21m/s^2

Generally the equation for Frictional force F_f is mathematically given as

F_f=m*a

F_f=m*(g-0.21)

F_f=87.5*(9.81-0.21)

Therefore

F_f=840N

6 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
3 years ago
What is the internal energy of a system including the kinetic and potential energy of its particles?
Gekata [30.6K]

Internal energy of a system is the product of specific heat and temperature included with number of moles. (i=nCT) and kinetic energy is product of Boltzmann constant and temperature.

  • All systems have a certain amount of energy that can be converted into other energy to do work. The kinetic energy within the molecules and atoms that make up the body and the energy generated by the intermolecular forces between them are collectively called internal energy.
  • In other words, the energy hidden in the system that can appear under different conditions can be called the internal energy of the system.

Internal energy is the sum of two energies:

(a) Thermal energy, which is the kinetic energy of molecules in random motion, and

(b) Potential energy of atoms.

  • Potential atomic energy arises from the atomic forces acting between the atoms of molecules and the intermolecular forces between molecules.
  • Total internal energy, E = kinetic energy (K.E.) + potential energy (P.E.).
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  • It is called the state function because the internal energy U is defined by the quantity that determines the state of the system at equilibrium. It is completely determined by the initial and final state of the system.
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8 0
11 months ago
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