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goblinko [34]
3 years ago
13

What is momentum in physics​

Physics
2 answers:
Otrada [13]3 years ago
8 0

Answer:

1. The quantity of motion of a moving body, measured as a product of its mass and velocity

2. The impetus gained by a moving object

viva [34]3 years ago
6 0

Answer:

Explanation:

the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction in three-dimensional spacethe product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction in three-dimensional space

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A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Keith_Richards [23]

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}

I = \dfrac{0.39\times 0.8^2}{12}+\dfrac{0.02\times 0.8^2}{4}+\dfrac{0.05\times 0.8^2}{4}

I =0.032\ kg.m^2

using conservation of energy

\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}

\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}

\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}

\omega=2.71 \rad/s

we know,

v = r ω

v = \dfrac{L}{2} \times 2.71

v = \dfrac{0.8}{2} \times 2.71

v = 1.084 m/s

3 0
3 years ago
what is the necessary condition within the hydrogen atom for balmer line photons to be absorbed by the hydrogen atom
luda_lava [24]

Explanation :

When an electron jumps from one energy level to another, the energy of atom gets changed.

If a photon gets absorbed, the electron will move to higher energy levels and then fall back to the lower energy levels. Then each time a photon will be absorbed whose energy is given by difference between the initial and final energy levels i.e

In Balmer series, the transition is from higher energy levels to n = 2.

So, the necessary condition for Balmer series is that the electron should be at first excited state or n = 2 level as shown in figure.


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The practice of playing top hits several times a day on radio stations is called _____. blocking rotation circulation format dri
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The practice of playing top hits several times a day on radio stations is called rotation.
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Dry air will break down if the electric field exceeds about 3.0×106v/m. part a what amount of charge can be placed on a capacito
Genrish500 [490]
The solution for this problem is:The charge would be now equal to:(electric constant) multiplied by the (field strength) multiplied by the (area) so plugging in our values, will give us:8.85 * 10^-12 As / (V * m) * 3 * 10^6 V/m * 0.055 m^2 = 1.46 e-6 amperes would be the answer
7 0
3 years ago
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti
NeX [460]

Answer:

a)  T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} ) ,  b) T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c)  x = d - \frac{m_1}{m_2} \  \frac{L}{2d},  d)  m₂ = m₁  ( \frac{ L}{2d} -1)

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )  - g m₂ - g m₁

       T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = d - \frac{m_1}{m_2} \  \frac{L}{2d}

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         \frac{m_1}{m_2} \ (0.5L -d) = d

          \frac{m_1}{m_2} = \frac{ d}{0.5L-d}

          m₂ = m₁  \frac{0.5 L -d}{d}

          m₂ = m₁  ( \frac{ L}{2d} -1)

5 0
3 years ago
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