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bagirrra123 [75]
3 years ago
13

A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

aves with a momentum of +13 kg • m/s.
What is the magnitude of the final momentum of the bowling pin if it has a mass of 1.5 kg?
Physics
1 answer:
dangina [55]3 years ago
8 0

Momentum should be conserved. The momentum of both objects must balance with their initial and final momentum.

Let m1 and v1 be the mass and velocity of the bowling ball

And m2 and v2 be the mass and velocity of the bowling pin

(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

30 kg m/s + (1.5 kg)(0 m/s) = 13kg m/s + 1.5v2f

V2f = 11.33 m/s

<span>So the momentum = 1.5 kg(11.33 m/s) = 17 kg m/s</span>

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Vector of magnitude 15 is added to a vector of magnitude 25. The magnitude of this sum
loris [4]

Explanation:

Given that,

Magnitude of vector A, |A| = 15

Magnitude of vector B, |B| = 25

We need to find the magnitude of this sum.

The maximum sum of the resultant vector,

R_{max}=|A_1|+|A_2|\\\\=15+25\\\\=45

The minimum sum of the resultant vector,

R_{min}=|A_1|-|A_2|\\\\=15-25\\\\=-10

So, the magnitude of this sum either 45 or -10.

6 0
3 years ago
An ion in a mass spectrometer follows a semicircular path of radius 14.8. What is the distance it travels?
aleksklad [387]

The circumference of a circle is (2π · the circle's radius).

The length of a semi-circle is  (1π · the circle's radius) =

                                                 (π · 14.8) = 46.5 (rounded)

(The unit is the same as whatever the unit of the  14.8  is.)

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3 years ago
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Juan makes an adjustment to an electromagnet that causes the electromagnet to lose some of its strength. What did Juan most like
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Reducing the amount of loops will cause a loss of strength, as the loops make the magnet stronger.
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4 years ago
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If a car goes down Lake at 30 miles per hour how far will it go in 0.25 hours?
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Explanation:

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7 0
3 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
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