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exis [7]
3 years ago
12

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas

t in neutral so as to just reach the top of the 150 m high hill at B with vB
Physics
1 answer:
Pavel [41]3 years ago
3 0

Complete question is:

A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.

Answer:

(V_A) = 31.32 m/s

Explanation:

We are given;

car's mass, m = 1200 kg

h_A = 100 m

h_B = 150 m

v_B = 0 m/s

From law of conservation of energy,

the distance from point A to B is;

h = 150m - 100 m = 50 m

From Newton's equations of motion;

v² = u² + 2gh

Thus;

(V_B)² = (V_A)² + (-2gh)

(negative next to g because it's going against gravity)

Thus;

(V_B)² = (V_A)² - (2gh)

Plugging in the relevant values;

0² = (V_A)² - 2(9.81 × 50)

(V_A) = √981

(V_A) = 31.32 m/s

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A charge of 12c passes through an electroplating apparatus in 2 min, what is the average current? ​
ryzh [129]

Answer:

Explanation:

If I am not wrong

current = charge/time

All you have to take care of is the units should be in the same system

so

current = 12/(2*60)         ---------  2 min = 2*60 sec

current  = 12/120  = 0.1 amp

8 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

     

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3 years ago
Consider five atoms from the second period : lithium , beryllium , boron , carbon , and nitrogen , which of these elements has t
kap26 [50]

Answer:

Lithium has the lowest electronegativity.

Explanation:

Electronegativity measures the tendency of an atom to attract the bonding pair of electrons. As we move left to right in a period, the number of shell remains same but the number of electron increases(negative charge increases. Hence, electronegativity also increases.

As Lithium is the left most element in this period. It has the lowest electronegativity value which is equal to 0.98.

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3 years ago
Which si unit whould be most appropriate for describing the length of a shoe?
aniked [119]

Answer:

Milimeters

Explanation:

This kind of question must be checked with everyday elements, in this way we will analyze each of the options until we reach the appropriate:

kilograms can not be because this unit serves to measure the mass not the length.

kilometers (km) is a unit of length, but however it does not serve for this measurement since a km is equivalent to 1000 meters, therefore measuring the length of a shoe with a ruler measuring 1000 meters is very impractical.

milliliters is a unit of volume, and can often be found in the measurement of volume content of liquids, for example 700 milliliters of milk or 1000 milliliters of water. Therefore it is impractical to measure a shoe with a unit of volume.

In this way the way to measure a shoe correctly is using a ruler or measuring instrument graduated in millimeters.

4 0
3 years ago
Which of the following is not a reason that coral reefs are important?
Brut [27]
My Answer: They contain a high diversity of organisms.

Hope I helped! :D
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