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stealth61 [152]
3 years ago
10

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his

acceleration
Physics
1 answer:
pantera1 [17]3 years ago
5 0

\Large {{ \sf {Question :}}}

<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

\Large {{ \sf {Given :}}}

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

\Large {{ \sf {Formulae  :}}}

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3>a \:  =  \frac{v - u}{t}</h3><h3>\Large {{ \sf {Step-by-step explanation :}}}</h3>

a \:  =  \frac{v - u}{t}  \\ or \:  \: a =  \frac{(15 - 5)}{10} m \: s^{ - 2}  \\ or \:  \: a \:  =  \frac{10}{10}m \: s^{ - 2} \\ or \:  \: a = 1m \: s^{ - 2}

\Large {{ \sf {Answer :}}}

<h3>His acceleration is </h3><h3>1m \: s^{ - 2}</h3><h3 /><h3 />
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A ferry is travelling at 10 m/s relative to the river towards the south. The river is flowing at 2 m/s relative to
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Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0
lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

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3 years ago
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