Answer:
Explanation:
According to Newton's third law of motion, forces always act in equal but opposite pairs. Another way of saying this is for every action, there is an equal but opposite reaction. This means that when you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted. 1.True 2.falues 3.true 4. not really sure on this one
The gravitational potential energy will increase by 423.36 J
<h3>How to determine the potential energy at ground level</h3>
- Mass (m) = 72 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Height (h) = 0 m
- Potential energy at ground level (PE₁) =?
PE = mgh
PE₁ = 72 × 9.8 × 0
PE₁ = 0 J
<h3>How to determine the potential energy at 60 cm (0.6 m)</h3>
- Mass (m) = 72 kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Height (h) = 0.6 m
- Potential energy at 60 cm (0.6 m) (PE₂) =?
PE = mgh
PE₂ = 72 × 9.8 × 0.6
PE₂= 423.36 J
<h3>How to determine the change in potential energy </h3>
- Potential energy at ground level (PE₁) = 0 J
- Potential energy at 60 cm (0.6 m) (PE₂) = 423.36 J
- Change in potential energy =?
Change in potential energy = PE₂ - PE₁
Change in potential energy = 423.36 - 0
Change in potential energy = 423.36 J
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The magnification <u>is 31.16.</u>
Magnification is the process of increasing the apparent size of something rather than its physical size. This increase is quantified by a calculated number, also called the "factor". If this number is less than 1, it means size reduction, sometimes called size reduction or reduction.
u = -19.3
f = -18.7 cm.
m = f/f-u
= -18.7/-18.7 +19.3
<u>= 31.16</u>
The term magnification refers to the size of the image produced by the lens compared to the size of the object. For lenses: Magnification "m" is the ratio of image height to object height. The magnification of a lens is defined as the ratio of image height to object height. It is also given by image distance and object distance. equal to the ratio of image distance to object distance.
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Answer:
<h3>a.</h3>
- After it has traveled through 1 cm :
- After it has traveled through 2 cm :
<h3>b.</h3>
- After it has traveled through 1 cm :
- After it has traveled through 2 cm :
Explanation:
<h2>
a.</h2>
For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient 
the formula is:
where I is the intensity of the beam, 
is the incident intensity and x is the length of the material traveled.
For our problem, after travelling 1 cm:
After travelling 2 cm:
<h2>b</h2>
The optical density od is given by:
.
So, after travelling 1 cm:
After travelling 2 cm:
Answer:
The required IVP is;
u'' + 196u = 0
where;
u(0) = 0 and
u'(0) = -10
Explanation:
See the attached for explanation
