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ratelena [41]
3 years ago
12

Why do the graphs differ?​

Physics
1 answer:
melisa1 [442]3 years ago
3 0

Well first graph represents rectangular hyperbola

vu = c^2 ( c is constant)

AS 1/v + 1/u = 1/f

Take1/ f to be constant c

1/v = c - 1/u

it is of the form y = - x + k

Slope = -1 having intercept k as shown in fig 2

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Explain why you cannot use mass or volume alone to identify substances.
Alex17521 [72]
Objects can have the same mass (but different <span>compositions). Only mass or volume cannot tell you if the object is solid or vo</span>lumes) or same volume (but different masses) 

4 0
3 years ago
A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop
KengaRu [80]

#A

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

\\ \rm\longmapsto Force=ma

\\ \rm\longmapsto Force=4.4(-1.74)

\\ \rm\longmapsto Force=-7.656N

#B

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

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

\\ \rm\longmapsto u=v-at

\\ \rm\longmapsto u=0-(-1.74)(1.27)

\\ \rm\longmapsto u=1.74(1.27)

\\ \rm\longmapsto u=2.2m/s

4 0
2 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
Anni [7]

Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
The red light from a helium-neon laser has a wavelength of 721.4 nm in air. Find the speed, wavelength, and frequency of helium-
saveliy_v [14]

Answer:

(a) the speed of helium-neon laser light in air is 3 x 10⁸ m/s

     the wavelength of helium-neon laser light in air is 721.4 nm

     the frequency of helium-neon laser light in air is 415.86 THz

(b)  the speed of helium-neon laser light in water is 2.26 x 10⁸ m/s

     the wavelength of helium-neon laser light in water is  542.4nm

     the frequency of helium-neon laser light in water is    416.67THz

(c) the speed of helium-neon laser light in glass is 2 x 10⁸ m/s

    the wavelength of helium-neon laser light in glass is  480.9nm

    the frequency of helium-neon laser light in glass is  415.88THz

From the results above, it can be seen that speed of the light is directly proportional to its wavelength, while the frequency of the light remained fairly constant for the different media.

Explanation:

Part (a) the speed, wavelength, and frequency of helium-neon laser light in air

Given;

wavelength of helium-neon laser light in air, λ = 721.4 nm

speed of light in air, v = 3 x 10⁸ m/s

v = f λ

where;

f is the frequency of helium-neon laser light in air

f = \frac{v}{\lambda} = \frac{3*10^8}{721.4 *10^{-9}} =4.1586*10^{14} \ Hz

f = 415.86 THz

Part (b) the speed, wavelength, and frequency of helium-neon laser light in water

refractive index of water = 1.33

Refractive \ index \ of \ water =\frac{speed \ of \ light \ in \ air}{speed \ of \ light \ in \ water} = \frac{wavelength \ of \ light \ in \ air}{wavelength \ of \ light \ in \ water}

speed \ of \ light \ in \ water = \frac{speed \ of \ light \ in \ air}{Refractive \ index \ of \ water} \\\\speed \ of \ light \ in \ water = \frac{3*10^8}{1.33} = 2.26 *10^8 \ m/s

Again;

wavelength \ of \ light \ in \ water = \frac{wavelength \ of \ light \ in \ air}{Refractive \ index \ of \ water} \\\\wavelength \ of \ light \ in \ water = \frac{721.4 \ nm}{1.33} = 542.4 \ nm

f = \frac{v}{\lambda} = \frac{2.26*10^8}{542.4 *10^{-9}} =4.1667*10^{14} \ Hz

f = 416.67 THz

Part (c) the speed, wavelength, and frequency of helium-neon laser light in glass

Refractive index of glass = 1.5

speed \ of \ light \ in \ glass = \frac{speed \ of \ light \ in \ air}{Refractive \ index \ of \ glass} \\\\speed \ of \ light \ in \ glass = \frac{3*10^8}{1.5} = 2 *10^8 \ m/s

Also;

wavelength \ of \ light \ in \ glass = \frac{wavelength \ of \ light \ in \ air}{Refractive \ index \ of \ glass} \\\\wavelength \ of \ light \ in \ glass = \frac{721.4 \ nm }{1.5} = 480.9 \ nm

f = \frac{v}{\lambda} = \frac{2*10^8}{480.9 *10^{-9}} =4.1588*10^{14} \ Hz

f = 415.88 THz

5 0
3 years ago
Read 2 more answers
Just number 9 I need help !!!!
Deffense [45]
Magnesium is in group 2 while sulfur is in group 6.

the group number determines the number of valence electrons.

Metals gives lose electrons (mg)
Non Metals gain electrons (s)

answer: Magnesium gives 2 electrons to sulfur.

Mg is now 2.8
Sulfur is 2.8.8

cuz ionic bonding
8 0
3 years ago
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