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3 years ago

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A 2 kg marble moving at 4 mi./s collides into a 1 kg marble at rest. After collision, the 2 kg marble speed decreased to 2 mi./s

. Calculate the velocity and speed of the 1 kg marble immediately after colliding.
Please show work and please help me as fast as possible because this is time

Please help as soon as possible please

Please I’m begging

1 answer:

Klio2033 [76]3 years ago

6 0

Answer:

$2\sqrt{6}$ $\frac{mi}{s}$

Explanation:

Assuming there is no waste of energy:

$K_{1} = K_{2}\\\frac{1}{2}m_{1}v_{1_{1}}^{2} + \frac{1}{2}m_{2}v_{2_{1}}^2 = \frac{1}{2}m_{1}v_{1_{2}}^{2} + \frac{1}{2}m_{2}v_{2_{2}}^2\\\\=> m_{1}v_{1_{1}}^{2} + m_{2}v_{2_{1}}^2 = m_{1}v_{1_{2}}^{2} + m_{2}v_{2_{2}}^2\\\\m_{1} = 2 kg, m_{2} = 1 kg, v_{1_{1}} = 4 \frac{mi}{s} , v_{2_{1}} = 0\\=> 32 = 8 + v_{2_{2}}^{2} => v_{2_{2}} = 2\sqrt{6} \frac{mi}{s}$

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two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram

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All atoms of the element potassium have 19 protons. One of the most stable types of potassium atoms has the mass number 39. How

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The formular for determinatio of mass number is this:
Mass number = number of proton + number of neutron
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3 years ago

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A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

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Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

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3 years ago

If an earthquake has a velocity of 41 m/s and a wavelength of 225 meters.

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Answer:

Frequency = 5.49Hz

Explanation:

Given the following data

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Making frequency the subject of formula, we have;

$Frequency = \frac {wavelength}{velocity}$

Substituting into the equation, we have;

$Frequency = \frac {225}{41}$

Frequency = 5.49Hz

Therefore, the frequency of the earthquake is 5.49 Hertz.

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3 years ago

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The plane's heading is 4.61 degrees south of east, which is the direction the pilot needs to steer the plane in order to reach her objective and the trip takes 1.83 hours.

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Distance is a numerical representation of the distance between two objects or locations.

The distance can refer to a physical length or an estimate based on other factors in physics or common use. |AB| is a symbol for the distance between two points A and B.

Let the pilot's heading be theta south with an east direction. The resultant the velocity will be;

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Hence the direction of the plane heading is 4.61 degrees south of east

b)

The resultant speed of the plane(V)is found as;

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Hence, the trip takes 1.83 hours.

To learn more about the distance refer to the link;

brainly.com/question/26711747

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2 years ago