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3 years ago

Do heavier bowling balls go faster than lighter ones?

Physics

2 answers:

cricket20 [7]3 years ago

8 0

Answer:

lighter ones will go faster

Explanation:

assuming you throw both bolwing ball with 10N of force. one is 3kg and the other is 5kg.

by using the equation: F = ma

a = F/m

a = 10/5 = 2m/s^2

a = 10/3 = 3.33333 m/s^2

Effectus [21]3 years ago

7 0

Piper rockelle and I just got off the phone number

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Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

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or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago

(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas

aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = $\frac{k*q}{r}$

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = $\frac{k*q1}{x}$ + $\frac{k*q2}{(1.63m-x)}$ = 0

⇒ $k*q1* (1.63m - x) = -k*q2*x$ :

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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3 years ago

Read 2 more answers

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Answer:

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3 years ago

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4 years ago

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PilotLPTM [1.2K]

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