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sukhopar [10]
3 years ago
10

(1) What distance does light travel in water during the time that it travels 1.10 m in vacuum?

Physics
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

1.   d = 8,270 10⁻¹ m , 2.  d = 7.33 10⁻¹ m , 3.  d = 5.116 10⁻¹  m

Explanation:

The speed of light in a vacuum is

         c = 2,998 10⁸ m / s

This speed is reduced in the material media, by interaction with the particles, the speed in the medium can be found by the index of refraction

          n = c / v

          v = c / n

As the speed is constant we can use the kinematic relationship

           v = d / t

           d = v t

           d = c t / n

We find the time it takes to travel the distance in a vacuum

            t = d / c

            t = 1.10 / 2.998 10⁸

            t = 3.669 10⁻⁹ s

1 in water refractive index is

            n = 1.33

            d = 2,998 10⁸ 3,669 10⁻⁹ / 1,333

            d = 8,270 10⁻¹ m

2 glass with n = 1.50

             d = 2,998 10⁸ 3,669 10⁻⁹ /1.5

             d = 7.33 10⁻¹ m

3 cubic zirconium n = 2.15

          d = 2,998 10⁸ 3,669 10⁻⁹ /2.15

          d = 5.116 10⁻¹  m

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Answer: let north be+ and south be-

therefore, -1 km  is the resultant displacement of the bird

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8 0
3 years ago
A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo
Inga [223]
The answer is 15.0N its explained in newtons third law hope this helps:)

4 0
3 years ago
Read 2 more answers
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
Advocard [28]

Answer:

a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

6 0
4 years ago
A 5 μF capacitor is connected to a 12 V battery. The charge on each plate of the capacitor is:
ziro4ka [17]
1 farad = 1 coulomb/volt

5 μF = 5 x 10⁻⁶ coulomb/volt

        = 60 x 10⁻⁶ coulomb / 12 volts

The charge is 60 x 10⁻⁶ coulombs = 6 x 10⁻⁵  (choice-A) 


3 0
4 years ago
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