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sammy [17]
3 years ago
13

A ball rolls at a speed of 6cm/s how far does the ball roll in 40 seconds ? ____cm

Physics
1 answer:
konstantin123 [22]3 years ago
6 0
240 cm you do 6x 40 and should get 240cm
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In order to walk barefoot on hot coals without hurting your feet
siniylev [52]

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5 0
3 years ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

6 0
3 years ago
Someone please help with this
SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

\displaystyle F=k\frac{q_1q_2}{d^2}

Where:

k=9\cdot 10^9\ N.m^2/c^2

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}

Factoring out 2/3:

\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}

Substituting the original force:

F'=\frac{2}{3}F

The new force is 2/3 of the original force

7 0
3 years ago
What is the acceleration of a 10kg pushed by a 5n force
Semenov [28]
Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a
8 0
3 years ago
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