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3 years ago

6

A bat hits a baseball with a force of 500N. The magnitude of the force that the baseball exerts on the bat is...

1 answer:

g100num [7]3 years ago

5 0

Answer:

45

Explanation:

because

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

A train traveled 85 mph for 7 hours how far did it travel?

Sergio [31]

Answer:

12.1429 miles per hour

Explanation:

4 0

3 years ago

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If the voltage drop across the first resistor is 13.00V, then how many volts remain for the 2nd resistor?

Scorpion4ik [409]

Answer:

2+1

Explanation:

2+1

8 0

2 years ago

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Find the mass of an object if a 40 N of force causes the object to accelerate at 5.5 m/s/s

Murrr4er [49]

Answer:

<h3>The answer is 8 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{40}{5} \\$

We have the final answer as

<h3>8 kg</h3>

Hope this helps you

7 0

3 years ago

How long do molecules of groundwater stay in the ground?

Brilliant_brown [7]

CORRECT ANSWER:

d. Anywhere from days to thousands of years.

STEP-BY-STEP EXPLANATION:

The whole question from book is

How long do molecules of groundwater stay in the

ground?

a. Days

b. Weeks

c. Months

d. Anywhere from days to thousands of years

4 0

3 years ago

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