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Natasha_Volkova [10]
3 years ago
11

A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

ection at 15.0 cm/s. after the collision, the 10.0 g marble moves to the right at 22.1 cm/s. find the velocity of the 25.0 g marble after the collision.
Physics
1 answer:
vovikov84 [41]3 years ago
6 0
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
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4 0
2 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
MatroZZZ [7]

Answer:

The wavelength is \lambda_2 =  534 *10^{-9} \ m

Explanation:

From the question we are told that

   The wavelength of the first light is  \lambda _ 1 =  587 \ nm

    The order of the first light that is being considered is  m_1  =  10

     The order of the second light that is being considered is  m_2  =  11

Generally the distance between the fringes for the first light is mathematically represented as

      y_1 =  \frac{ m_1  * \lambda_1 *  D}{d}

 Here  D is the distance from the screen

 and    d  is the distance of separation of the slit.

      For the second light the distance between the fringes is  mathematically represented as

         y_2 =  \frac{ m_2  * \lambda_2 *  D}{d}

Now given that both of the light are passed through the same double slit

       \frac{y_1}{y_2}  =  \frac{\frac{m_1 *  \lambda_1 * D}{d}  }{\frac{m_2 *  \lambda_2 * D}{d}  } = 1

=>    \frac{ m_1 *  \lambda _1  }{ m_2  *  \lambda_2} =  1

=>     \lambda_2 =  \frac{m_1 *  \lambda_1}{m_2}

=>    \lambda_2 =  \frac{10  *   587 *10^{-9}}{11}

=>   \lambda_2 =  534 *10^{-9} \ m

4 0
3 years ago
A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on
n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0
2 years ago
A 0.45-m metal rod moves 0.11 m in a direction that is perpendicular to a 0.80-T magnetic field in an elapsed time of 0.036 s. A
Gwar [14]

Answer:

1.1 V

Explanation:

L = 0.45 m

d = 0.11 m

B = 0.80 T

t = 0.036 s

Let e be the emf.

e = B v L

e = 0.80 x 0.11 x 0.45 / 0.036 = 1.1 V

3 0
3 years ago
Read 2 more answers
Which type of force absorbs shock in vehicles
Licemer1 [7]

Answer:

Vehicles typically employ both hydraulic shock absorbers and springs or torsion bars. In this combination, "shock absorber" refers specifically to the hydraulic piston that absorbs and dissipates vibration.

Explanation:

hope this helps

5 0
3 years ago
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