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madam [21]
3 years ago
8

Rt c if you were to hit a ping pong ball of mass 0.0029 g with the same force that caused a 0.058-g tennis ball to move with an

acceleration of 10 m/s2 , what would the acceleration a of the ping pong ball be? express your answer in meters per second squared.

Physics
2 answers:
kati45 [8]3 years ago
7 0
F=ma
You know the acceleration and mass of the tennis ball, now you can find the force applied by substituting it into the formula. Also I am assuming you meant to write your units in kg because that seems very light for grams. I am sorry if I misunderstood. 
F=ma
  = (0.058kg)(10m/s²)
  = 0.58N
Now that you know how much force was exerted you can solve for how much acceleration the ping pong ball will have.
F=ma
a=F/m
  =(0.58N)/(0.029kg)
  =20m/s²
Hope this helps!
Daniel [21]3 years ago
5 0

<em>The acceleration of the ping pong ball would be </em><em>200 m/s²</em>

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\boxed {\Sigma F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

mass of ping pong ball = m₁ = 0.0029 g

mass of tennis ball = m₂ = 0.058 g

acceleration of tennis ball = a₂ = 10 m/s²

<u>Asked:</u>

acceleration of ping pong ball = a₁ = ?

<u>Solution:</u>

\Sigma F_{\texttt{ping pong ball}} = \Sigma F_{\texttt{tennis ball}}

m_1 a_1 = m_2 a_2

0.0029 \times a_1 = 0.058 \times 10

0.0029 \times a_1 = 0.58

a_1 = 0.58 \div 0.0029

a_1 = 200 \texttt{ m/s}^2

\texttt{ }

<h3>Conclusion :</h3>

<em>The acceleration of the ping pong ball would be </em><em>200 m/s²</em>

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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Answer:

0.1 mm

Explanation:

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Hope This Helped

3 0
1 year ago
3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?
muminat

Answer:

7 meters, 2.8 meters

Explanation:

work done (nm) = force (n) * distance (m)

140= 20 * m

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4 0
2 years ago
Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t
lorasvet [3.4K]

Answer:

V_3\approx 4.28\,\,V

I_1=0.0572\,\,amps

I_3\approx 0.171\,\,amps

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega

By knowing this, we can estimate the total current through the circuit,:

Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps

So approximately 0.17  amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V

So now we know that the potential drop across the parellel resistors must be:

10 V -  4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps

4 0
3 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
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MatroZZZ [7]
It would take 500 seconds, because 2500 divided by 5 is 500
3 0
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