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dimaraw [331]
3 years ago
7

Two cars are raised to the same elevation on service- station lifts. If one car is twice as massive as the other, how do their p

otential energies compare?
Physics
1 answer:
Brums [2.3K]3 years ago
7 0

Answer:

The potential energy of the more massive one is twice that of the other.

Explanation:

Potential energy is given by

<em>PE</em> = <em>mgh</em>

where <em>m</em> = mass of body, <em>g</em> = acceleration of gravity and <em>h</em> = height or elevation.

For the less massive car, let the mass be m_1. Then its <em>PE</em> is

PE_1 = m_1gh

For the massive car, let the mass be m_2.  Its <em>PE</em> is

PE_2 = m_2gh

But m_2 =2m_1

\therefore PE_2 = 2m_1gh = 2(m_1gh) = 2PE_1

Hence, the potential energy of the more massive one is twice that of the other.

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A person is lying on a diving board 3.50 m above the surface of the water in a swimming pool. She looks at a penny that is on th
sergij07 [2.7K]

Answer:

1.995 m

Explanation:

Distance of penny as seen by the person = 5 m

Height of person from water surface = 3.50 m

Apparent depth of penny = 5 - 3.50 = 1.5 m

refractive index of water, n = 1.33

real depth / apparent depth = n

real depth = 1.33 x 1.5 = 1.995 m

Thus, the actual depth of water at that point is 1.995 m.

4 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 pie is cooked in an oven at 200 °C. The aluminium film that ckvered the pie can be touched soon after it is removed while the
GREYUIT [131]

Answer and Explanation:

The aluminum is more productive in the absorption and heat transfer to other particles. It instantly converts heat absorbed from the environment into the atmosphere when removed from the oven, enabling us to operate with it faster than the pie that takes much longer to convert heat to the environment.

So this is the reason for pie to be the dangerously hot

8 0
3 years ago
I need help
musickatia [10]

Answer:

C

Explanation:

4 0
3 years ago
A student is trying to calculate the density of an ice cream cone. She already knows the mass, but she needs to determine the vo
vladimir2022 [97]
V= 1/3 π r²h 
this is the formula for a cone hope this helps :) 

5 0
3 years ago
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