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Serggg [28]
2 years ago
8

Of the 5000 species of mammals, 250 species are carnivorous. What is the ratio of carnivorous to mammals?

Physics
1 answer:
Mekhanik [1.2K]2 years ago
4 0

Answer: The ratio of carnivorous species to all mammal species is 1:20

Explanation:

250:5000 = 250/250 = 1

250:5000 = 5000/250 = 20

Answer = 1:20

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Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.

The velocity vector of the cat when it reaches the ground will be:

v = (v0x, v0y + g · t)

v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)

v = (44 m/s, -34.3 m/s)

The magintude of the vector "v" is calculated as follows:

|v| = \sqrt{(44 m/s)^{2}+(-34.3 m/s)^{2}} = 55.8 m/s

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

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When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the
KonstantinChe [14]

Answer:

The ratio is  KE : TM  =  0.75

Explanation:

from the question we are told that

  The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical  energy of the mass is mathematically represented as

        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

        TM  =  KE  + PE

=>     KE = TM  - PE

Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

      x = \frac{A}{2}

So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

=>    KE =  0.375  *  k  *  A^2

So the ratio of  KE :  TM is  mathematically represented as

       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

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An LED is useful because when a current passes through it, it gives out light.

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