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Phoenix [80]
3 years ago
11

This is the fundamental force which holds particles together to form protons and neutrons

Physics
2 answers:
Alex17521 [72]3 years ago
8 0
The four fundamental forces of nature are gravity, electromagnetism, the weak force, and the strong force. 
The strong force is what holds particles together to form protons and neutrons.  
Zinaida [17]3 years ago
3 0

The correct answer is a strong force

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Two balls, each with a mass of 0.890 kg, exert a gravitational force of 8.06 × 10−11 n on each other. how far apart are the ball
Bond [772]
This problem involves Newton's universal law of gravitation and the equation to follow would be.

F = GM₁M₂/r²  

Given: M₁ = 0.890 Kg;  M₂ = 0.890 Kg;  F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²

Solving for distance r = ?

r = √GM₁M₂/F

r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N

r = 0.81 m 
6 0
3 years ago
A tea kettle is boiling on the stove, and all of a sudden the whistle in the spout of the kettle starts shrieking. Steam rising
fenix001 [56]

Answer:

The steam represents evaporation.

Explanation:

4 0
2 years ago
Read 2 more answers
the other end of the pipe. With the aide of the pipe, does the applied force produce a smaller torque, a greater torque, or the
pav-90 [236]

Answer:

With the addition of the pipe we have a greater torque.

Explanation:

We need to complete the description of the problem, searchin in internet we have:

"Sometimes, even with a wrench, one cannot loosen a nut that is frozen tightly to a bolt. It is often possible to loosen the nut by slipping one end of a long pipe over the wrench handle and pushing at the other end of the pipe. With the aid of the pipe, does the applied force produce a smaller torque, a greater torque, or the same torque on the nut?"

With the addition of the pipe we have a greater torque, as it increases the distance or radius of torque.

We know that torque is defined, as the product of force by distance, in this way we have:

T = F * d

where:

T = torque [N*m]

F = force [N]

d = distance [m]

We can see in the above equation, that increasing the distance increases torque proportionally.

3 0
3 years ago
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
A student gathered two boxes of the same size made of different materials glass and clear plastic. She placed them on a windowsi
ch4aika [34]

Answer:

dependent variable

Explanation:

8 0
2 years ago
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