The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo
1 answer:
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
The mass of the baseball, m = 0.15 kg
The speed at which it moves, v = 30 m/s
Time at which the baseball was in contact with the bat, t = 0.75 ms
t = 0.75/1000 s
t = 0.00075 s
The impulsive force is given by the formula:
Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:
The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N
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The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.
<h3>What is Columb's law?</h3>The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.
They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.
The given data in the problem is;
q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C
F is the repulsive force = 36 mN =36 ×10⁶ N
d is the distance = 0.70 m
The Coulomb force is found as;
Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.
To learn more about Coulomb's law, refer to the link;
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