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AlladinOne [14]
3 years ago
5

WILL MARK BRAINIEST DO QUICKKKK ISTGG PLS ​

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
3 0

Answer:

the answer is up the bird will go up

Explanation:

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Which bright solar feature is shown in the picture above?
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Answer : (B) Prominence

Explanation :

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<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

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4 years ago
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the choices below each describe the appearance of an h-r diagram for a different star cluster. which cluster is most likely to b
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7 0
1 year ago
What is electricity
fgiga [73]
Electricity is a form of energy. Electricity is the flow of electrons. All matter is made up of atoms, and an atom has a center, called a nucleus. The negative charge of an electron is equal to the positive charge of a proton, and the number of electrons in an atom is usually equal to the number of protons. Hope this helped. ;)
7 0
3 years ago
One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.
saul85 [17]
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3 0
3 years ago
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Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c
Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is  v_A= 4m/s

Explanation:

From the question we are told that

    The length of the string is  L = 1m

     The initial speed of block A is u_A

     The final speed of block A is  v_A = \frac{1}{2}u_A

      The initial speed of block B is u_B = 0

      The mass of block A  is  m_A = 7kg  gh

      The mass of block B is  m_B  = 2 kg

According to the principle of conservation of momentum

       m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}

Since block B at initial is at rest

       m_A u_A  = m_Bv_B + m_A \frac{u_A}{2}

      m_A u_A  - m_A \frac{u_A}{2} = m_Bv_B

          m_A \frac{u_A}{2} = m_Bv_B

  making v_B the subject of the formula

             v_B =m_A \frac{u_A}{2 m_B}

Substituting values

               v_B =\frac{7 u_A}{4}  

This v__B is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity  of v__B' at  the top of the vertical circle  

 The angular centripetal acceleration  would be mathematically represented

                   a= \frac{v^2_{B}'}{L}

Note that  this acceleration would be toward the center of the circle

      Now the forces acting at the top of the circle can be represented mathematically as

         T + mg = m \frac{v^2_{B}'}{L}

    Where T is the tension on the string

  According to the law of energy conservation

The energy at  bottom of the vertical circle   =  The energy at the top of

                                                                                the vertical circle

   This can be mathematically represented as

                 \frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L

From above  

                (T + mg) L = m v^2_{B}'

Substitute this into above equation

             \frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L  + mg 2L  

             \frac{49 mv_A^2}{16}  = \frac{1}{2} (T + mg) L + mg 2L

          \frac{49 mv_A^2}{16}  = T + 5mgL

The  value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is  zero

        This is mathematically represented as

                      \frac{49 mv_A^2}{16}  = 5mgL

making  v_A the subject

            v_A = \sqrt{\frac{80mgL}{49m} }

substituting values

          v_A = \sqrt{\frac{80* 9.8 *1}{49} }

              v_A= 4m/s

     

6 0
3 years ago
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