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denis23 [38]
1 year ago
7

What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

electric field of magnitude 590 N/C
Physics
1 answer:
Pani-rosa [81]1 year ago
8 0

     charge must be equal to 5.74 ×10⁻⁵

 In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

     →    Fnet =0

     →    mg =  qE

 substituting the values we get :

         0.00345 × 9.81 =  q × 590

   →       q = 5.74 ×10⁻⁵

    Hence the charge must be equal to   5.74 ×10⁻⁵.

   Learn more about charges here:

          brainly.com/question/26092261

                    # SPJ4

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20 Yards

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Start      End

Total displacement(D)  20 yards (East).

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How is a magnetable to pick up bits of metal without actually touching them?
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C.There is an invisible magnetic field around the magnet where it exerts magnetic force

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What does the left y-axis show?
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The correct option is;

a- sea surface temperature anomaly, in degrees Celsius

Explanation:

From the diagram related to the question we have two graphs super imposed of Sea surface temperature anomaly, in degrees Celsius and cholera incidence anomaly (%) both plotted against time in years.

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2 years ago
Question Part Points Submissions Used A pitcher throws a 0.200 kg ball so that its speed is 19.0 m/s and angle is 40.0° below th
postnew [5]

Answer:

The impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

Explanation:

In a problem of impulse and shocks we must use the impulse equation

       I = dp = pf-p₀         (1)

       p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

      V₀ = -19 m / s

      θ₀ = 40.0º  

      Vf = 46.0 m / s

      θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

      Vcx = Vo cos θ

      Voy = Vo sin θ

      Vox= -19 cos (40) = -14.6 m/s

      Voy = -19 sin (40) =  -12.2 m/s

      Vfx = 46 cos 30 = 39.8 m/s

      Vfy = 46 sin 30 =  23.0 m/s

   a) We already have all the data, substitute and calculate the impulse for each axis

      Ix = pfx -pfy

      Ix = m ( vfx -Vox)

      Ix = 0.200 ( 39.8 – (-14.6))

      Ix = 10.88 N s

      Iy = m (Vfy -Voy)

      Iy = 0.200 ( 23.0- (-12.2))

      Iy=  7.04 N s

In vector form it remains

       I =  (10.88 i^ + 7.04 j^) N s

   b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

         I = Fpro Δt

In the first interval

        Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

         Fpro = (Fm +0) / 2

In the second interval the force is constant

          Fpro = Fm

In the third interval

         Fpro = (0 + Fm) / 2

Let's replace and calculate

         I =  Fpro1 t1 +Fpro2 t2  +Fpro3 t3

         I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³  

         I = Fm  24 10⁻³ N s

         Fm = I / 24 10⁻³

         Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

         Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

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