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AlekseyPX
3 years ago
11

Which statement about the positive and negative value of speed compared to velocity is true?

Physics
1 answer:
mestny [16]3 years ago
3 0
D is the answer
Velocity maybe negative or positive
while speed is always positive
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Flint glass materials ​
Kipish [7]

Answer:

Flint glass is combination of silicon dioxide (SiO2) with lead or potassium. It creates a relatively high refractive index and high degree of light dispersing power compared to other types of glass.

Explanation:

:)

7 0
3 years ago
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What is it that lives if it is fed, and dies if you give it a drink?
Anna71 [15]

Fire is it that lives if it is fed, and dies if you give it a drink.

<h3><u>Explanation:</u></h3>

Fire is very essential part of human life. It is used for cooking food and for other important activities. Without fire we cannot not survive. Something or the other should be heated before consumption and this can be achieved only with fir. It is also used in the darker places for viewing many things around us.

Thus, fire can survive if we give fuel or any wooden pieces and when water is poured on it it will turn off. Hence Fire is the one that survives when it is fed and dies when water is given as a drink to it.

4 0
3 years ago
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capp
Bond [772]

Answer:

= 925.92 N

≅ 926N

Explanation:

Pressure due to car = pressure due to applied force  

12000/18^2 = Force / 5^2

force = 12000 * 25/ 324

= 925.92 N

For equilibrium

Pressure1 = Pressure2

A1F1 = A2F2

12000*pi*(5^2) = F2 ( pi)*(18^2)

so, F2 = Applied force to lift car = 925.92 N

Pascal's principle

Pressure1 = Pressure2

F1/A1 = F2/A2 (F=force and A=area)

A1 =Pi*(0.05)²

A2 =Pi(0.18)²

F2=12000

F1 = 12000*(0.05)² / (0.18)² = 926N

7 0
3 years ago
An object at rest cannot remain at rest unless which of the following holds?View Available Hint(s)An object at rest cannot remai
Leni [432]

Answer:

True The net force must be zero for the acceleration to be zero

Explanation:

In order to analyze the statements of this problem we propose your solution.

First let's look at Newton's first, which stable that every object is at rest or with constant speed unless something takes it out of this state (acceleration)

Now let's look at the second postulate, which says that force is related to the product of the mass of a body and its acceleration.

As a result of these two laws, for a body is a constant velocity the summation force on it must be zero.

Now we can analyze the statements given.

True The net force must be zero for the acceleration to be zero

False. If the force is different from zero, there is acceleration that changes the speeds

False. There may be forces, but the sum of them must be zero

False. If a force acts, the acceleration is different from zero and the speed changes

5 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
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