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VashaNatasha [74]
3 years ago
5

When a physical change occurs, the mass of the substance is conserved. This means that the total mass of the substance remains t

he same from beginning to end. The physical properties of the substance, such as size and shape, may change, but the amount of matter in the substance does not change.
Physics
1 answer:
Doss [256]3 years ago
7 0
Yes yes ! How right you are ! Truer words were never spoken.
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What is the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosp
sineoko [7]

The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

To find the answer, we have to know about the pressure.

<h3>How to find the weight of a column of air?</h3>
  • As we know that the expression of pressure as,

                 P=\frac{F}{A}

where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.

  • It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,

             P=1atm=1.013*10^5Pascals

  • From this, the value of weight will be,

            F=mg=P*A=1.013*10^5*4.5=4.56*10^5N

Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.

Learn more about the pressure here:

brainly.com/question/12830237

#SPJ4

4 0
1 year ago
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 475 turns, its
VMariaS [17]

Answer:

Output voltage is 1.92 volts.

Explanation:

Given that,

Number of turns in primary coil, N_p=475

Number of turns in secondary coil, N_s=8

Input voltage, V_i=114\ V

We need to find the voltage output of a transformer used for rechargeable flashlight batteries. For a transformer, the number of turns and the voltage ratio is given by :

\dfrac{N_p}{N_s}=\dfrac{V_i}{V_o}\\\\\\V_o=114\cdot\dfrac{8}{475}\\\\V_o=1.92\ V

So, the output voltage is 1.92 volts.

3 0
3 years ago
An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east. Immediate
Fantom [35]

Answer:

6.25 m/s

Explanation:

mass of man (m1) = 80 kg

mass of boy (m2) = 20 kg

mass of man and boy after collision (m12)= 20 + 80 = 100 kg

velocity of man and boy after collision (v) = 2.5 m/s

angle θ = 60 °

How fast was the boy moving just before the collision ?

  • From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
  • M₁₂ =  total momentum after collision = m12 x v = 100 x 2.5 = 250
  • Mboy = momentum of the boy before collision = m2 x Velocity of boy
  • Mman = momentum of the man before collision = m1 x velocity of man  
  • from the triangle, cos θ = \frac{Mboy}{M₁₂}

        cos 60 = \frac{Mboy}{250}

        Mboy = 250 x cos 60 = 125

  • recall that momentum of the boy (Mboy) also = m2 x Velocity of boy

        therefore

        125 = 20 x velocity of boy

         velocity of boy = 125 / 20 = 6.25 m/s

4 0
3 years ago
A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the t
Virty [35]

Answer:

B)   d = √  ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

          Em₀ = U = m g h₁

Final point. Lower (slide bottom)

            Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

             mgh₁ = ½ m v² + mgh₂

             v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

           x = v₀ₓ t

          y = v_{oy} t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

         t = √ 2h₂ / g

We substitute in the other equation

         d = √ (2g (h₁-h₂))  √ 2h₂ / g

         d = √ (4 h₂ (h₁-h₂))

         H = h₁ + h₂

         h₁ = H -h₂

         d = √  ( 4 h₂ ( H - 2h₂))

3 0
3 years ago
Read 2 more answers
Which of the following is an example of velocity?
pav-90 [236]
C :) cause a direction has nothing do do with velocity !
8 0
3 years ago
Read 2 more answers
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