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harina [27]
4 years ago
13

A baseball catcher is performing a stunt for a television commercial. He will catch a baseball (mass 145 g) dropped from a heigh

t of 66.0 m above his glove. His glove stops the ball in 0.0118 s. What is the force (in N) exerted by his glove on the ball? (Indicate the direction with the sign of your answer. Assume the baseball is traveling in the negative direction.)
Physics
1 answer:
Tomtit [17]4 years ago
4 0

Answer:

Explanation:

The ball is going down with velocity. It must have momentum . It is stopped by

catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball  by the gloves  .

The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .

Let us measure the force applied on the ball .

Final velocity after the fall by 66 m

V = √ 2gh

= √ 2x9.8 x 66

35.97 m /s

Momentum = m v

0.145 x 35.97

= 5.2156 kgms⁻¹

Change in momentum

= 5.2156 - 0

= 5.2156

Rate of change of momentum

=  Change of momentum / time = 5.2156 / .0118

= 442 N

if R be the force exerted by gloves to stop the ball

R - mg represents the net force which stops the ball so

R - mg = 442

R = 442 + mg

= 442 + .145 x 908

443.43 N

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The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

\vec r_{1} = (-0.500, 0)\,[m]

\vec r_{2} = (+0.500, 0)\,[m]

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Then, distances of the former two particles particles respect to the latter one are found now:

\vec r_{13} = (+0.500, +0.500)\,[m]

r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}

r_{13} =\frac{\sqrt{2}}{2}\,m

\vec r_{23} = (-0.500, +0.500)\,[m]

r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}

r_{23} =\frac{\sqrt{2}}{2}\,m

The resultant force is found by Coulomb's law and principle of superposition:

\vec R = \vec F_{13}+\vec F_{23} (1)

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\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13}  +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23} (2)

Where:

  • k - Electrostatic constant, in newton-square meters per square Coulomb.
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If we know that k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}, q_{1} = 2\times 10^{-6}\,C, q_{2} = 2\times 10^{-6}\,C, q_{3} = 2\times 10^{-6}\,C, r_{13} =\frac{\sqrt{2}}{2}\,m, r_{23} =\frac{\sqrt{2}}{2}\,m, \vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}  \right) and \vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right), then the vector force on charge q_{3} is:

\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}  \right)

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\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]

And the magnitude of the <em>electrical</em> force on charge q_{3} (R), in newtons, due to the others is found by Pythagorean theorem:

R = 0.102\,N

The magnitude of <em>electrical</em> force on charge q_{3} due to the others is 0.102 newtons. \blacksquare

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(a) 0.74 m/s^2

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The component of the weight parallel to the inclined plane is:

W= m g sin \theta

where m is the mass of the skier, g=9.81 m/s^2 and \theta=10^{\circ}.

The frictional force is instead

F_f = -\mu m g cos \theta

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