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Svet_ta [14]
3 years ago
14

How to find the area of the side with the maximum surface area of a rectangle

Physics
2 answers:
Aleks04 [339]3 years ago
8 0

Answer:

Let one side = Let the other side= Let area = Let perimeter= Now substitute into area formula

Explanation:

Fed [463]3 years ago
8 0

Answer:

Let one side Let the other side Let area  Let perimeter

Explanation:

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Which one describes how much matter an object contains, using SI units?
ahrayia [7]
Si units or Systeme' de Internationale' is a widely adopted unit system in measuring basic and derived dimensions In this case, the SI units here are kilograms, meter and seconds. Pounds is an English unit. mass is the measure of <span>how much matter an object contains, hence the answer is A. 43 kg.</span>
5 0
3 years ago
Read 2 more answers
Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated
Artemon [7]

Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>
  • Gravitation is the force of attraction between any two bodies.
  • Every body in the universe attracts every other body with a force.
  • This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
  • Mathematically we can expressed it as,

                        F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2

<h3>How to solve the problem?</h3>
  • Here, we have given with the data's,

                      M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m

  • Thus, the force of attraction between these two bodies will be,

               F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

Learn more about the Newton's law of gravitation here:

brainly.com/question/28045318

#SPJ4

6 0
2 years ago
Geologists use tools other than the law of superposition to help them with relative dating. Which statements describe those tool
White raven [17]

Answer: Cross-cutting features are always younger than the surrounding rock.

When material erodes before sediment is deposited on it, a geologic gap results.

Explanation:

The options include:

1. An unconformity is created when lava pours out on Earth’s surface.

2. Faults are the result of volcanic activity.

3. Intrusions and extrusions are sedimentary formations.

4. Cross-cutting features are always younger than the surrounding rock.

5. When material erodes before sediment is deposited on it, a geologic gap results.

The law of superposition simply states that when there is a layers of rocks, we would see that the younger layers will lie and be on top of the layers that are older.

Other tools that can help scientist with relative dating are:

• Cross-cutting features are always younger than the surrounding rock.

• When material erodes before sediment is deposited on it, a geologic gap results.

6 0
3 years ago
During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common iso
mariarad [96]

Answer:

21.55 m

Explanation:

6 0
3 years ago
An Airbus A380 airliner can takeoff when its speed reaches 80 m/s. Suppose its engines together can produce an acceleration of 3
Tomtit [17]

Answer:

(a). The time is 26.67 sec.

(b). The distance traveled during this period is 1066.9 m.

Explanation:

Given that,

Speed = 80 m/s

Acceleration = 3 m/s

Initial velocity = 0

(a). We need to calculate the time

Using equation of motion

v = u+at

t = \dfrac{v-u}{a}

Put the value into the formula

t = \dfrac{80-0}{3}

t =26.67\ sec

The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times3\times(26.67)^2

s =1066.9\ m

The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

7 0
3 years ago
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