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ladessa [460]
3 years ago
9

A pitcher throws a softball toward home plate. When the ball hits the catcher’s mitt, its horizontal velocity is 32 meters/secon

d. The softball’s velocity goes to 0 meters/second in 0.8 seconds when caught. If the softball has a mass of 0.2 kilograms, what’s the force of the impact? Use F = ma, where . The force of the impact is newtons.
Physics
2 answers:
il63 [147K]3 years ago
6 0

Answer:  I just took the test on Plato and the correct answer is 8

Explanation:

To calculate the force of impact F, first lets calculate the acceleration a of the ball:

a=v/t where v is the velocity of the ball and t is time

a=32/0.8=40 m/s²

To get the force F we need the Newtons second law:

F=m*a where m is the mass of the ball and a is the acceleration.

F=m*a= 0.2*40 = 8 N

So the impact force is F= 8 N.

Read more on Brainly.com - brainly.com/question/3940090#readmore

n200080 [17]3 years ago
5 0
To calculate the force of impact F, first lets calculate the acceleration a of the ball: 

a=v/t where v is the velocity of the ball and t is time

a=32/0.8=40 m/s²

To get the force F we need the Newtons second law:

F=m*a where m is the mass of the ball and a is the acceleration.

F=m*a= 0.2*40 = 8 N

So the impact force is F= 8 N.
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bagirrra123 [75]

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

  • q is magnitude of master charge
  • r is distance of separation
  • k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

  • E is electric field
  • q is the test charge

F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

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8 0
1 year ago
Which statement is a hypothesis?
olga_2 [115]
B, do earthworms prefer bright light or darkness!
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2 years ago
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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
Marrrta [24]

Answer:

a

 Solid Wire     I  =   0.01237 \  A      

  Stranded  Wire  I_2  =   0.00978 \  A

b

  Solid Wire   R  = 0.0149 \ \Omega

   Stranded  Wire  R_1  = 0.0189 \ \Omega

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  r_1  = 1.53 mm = 0.0015 \  m

  The radius of  each strand is  r_0 =  0.306 \ mm =  0.000306 \ m

  The current density in both wires is  J  =  1750 \  A/m^2

Considering the first wire

     The  cross-sectional area of the first wire is

      A   = \pi  r^2

= >  A   = 3.142 *  (0.0015)^2

= >  A   = 7.0695 *10^{-6} \  m^2

Generally the current in the first wire is    

     I  =  J*A

=>  I  =  1750*7.0695 *10^{-6}

=>  I  =   0.01237 \  A

Considering the second wire  wire

The  cross-sectional area of the second wire is

     A_1  =  19 *  \pi r^2

=>     A_1  =  19 *3.142 *  (0.000306)^2

=>  A_1  =  5.5899 *10^{-6} \  m^2

Generally the current is  

      I_2  =  J  *  A_1

=>    I_2  =   1750  *  5.5899 *10^{-6}

=>    I_2  =   0.00978 \  A

Considering question two  

 From the question we are told that

     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

Generally the resistance of the first wire is mathematically represented as

    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

=> R_1  = 0.0189 \ \Omega

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3 years ago
If 20 beats are produced within one second, which of the following frequencies could possibly be held by two sound waves traveli
NeTakaya

Answer:

D. 22 Hz and 42 Hz

Explanation:

  • When two waves with different frequency travelling in the same medium meet each other, they produce an interference pattern called beat.
  • <em><u>The frequency of the beat produced is equivalent to </u></em><em><u>the difference between the individual frequencies of the two waves involved.</u></em>
  • <em><u>Therefore; in this case since the frequency of the beat is 20 Hz, that is from 20 beats per second.</u></em>
  • We need to find a pair from the choices whose frequency difference is 20 Hz.
  • This happens to be choice D. 22 Hz and 42 Hz,  that is 42 Hz - 22 Hz = 20 Hz
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Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

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          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

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