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Ulleksa [173]
2 years ago
7

To measure the current through and the voltage across a resistor in a circuit, you should place the ammeter in _____ with the re

sistor and the voltmeter in _____ with the resistor.
Physics
2 answers:
zysi [14]2 years ago
8 0

Answer:

  • ammeter: series
  • voltmeter: parallel

Explanation:

The current is the same through components connected in series. If you want to measure the current in a component, the ammeter must be placed in series.

__

The voltage is the same at the terminals of components connected in parallel. If you want to measure the voltage across a component, the voltmeter must be connected across the component, that is, in parallel with it.

egoroff_w [7]2 years ago
5 0

\huge \mathbb \pink{ANSWER:}

To measure the current through and the voltage across a resistor in a circuit, you should place the ammeter in <u>series</u> with the resistor and the voltmeter in <u>parallel</u> with the resistor.

#CarryOnLearning

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(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

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5 0
2 years ago
How to represent milligram in kilogram by standard formula?
Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take   \frac{1}{1,000,000} which equals 0.000001 kg

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

     

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3 years ago
A process occurs in which a system's potential energy increases while the environment does work on the system.
iogann1982 [59]
Ok, now what do you want to know about it?
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3 years ago
An autographed baseball rolls off of a 0.91 m high desk and strikes the floor 0.84 m away from the desk. How fast was it rolling
Rudik [331]
The initial velocity (its speed before falling off) is approximately 1.95 m/s
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3 years ago
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