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Bezzdna [24]
3 years ago
11

Describe a light wave and explain how light wave travel through solids liquids and gasses

Physics
1 answer:
marin [14]3 years ago
8 0
Through refraction , it bends as it passes into a solid object
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on touching electroscope gets positively charged, so answer is B. conduction

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3 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance  = 5 m

a) for friction less  

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh

             v^2 = u^2 - 2gh

             v = \sqrt{u^2- 2 g h cos 22^0}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }

                    v = 17.58 m/s

b) considering the friction

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl

             v^2 = u^2 - 2gh-2\mu_kmgl

             v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }

                   v = 17.16 m/s

7 0
3 years ago
Which of these statements is true?
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C.) Meiosis involves two cycles of cell division

Hope this helps!
6 0
3 years ago
Read 2 more answers
Jvhftfcticgvigybvhbjnjndjanvjisn
nirvana33 [79]

Answer:

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4 0
3 years ago
Read 2 more answers
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
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