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Alinara [238K]
3 years ago
13

You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A

t what speed does the baseball leave your hand?
Physics
1 answer:
garri49 [273]3 years ago
7 0

Answer:

The speed with which the baseball leaves the hand = 20.58 m/s

Explanation:

The time take to reach highest height during a projectile's flight is given by

t = (u sin θ)/g

u = initial velocity of the baseball = ?

θ = angle of throw above the horizontal

g = acceleration due to gravity = 9.8 m/s²

1.05 = (u sin 30)/9.8

u = (1.05 × 9.8)/0.5

u = 20.58 m/s

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Ainat [17]

The total electrostatic force on charge A is 28 \mu N

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

Here we have three positively charged particles A,B and C, located at the following positions:

x_A = 0\\x_B = 10 m\\x_C = 20 m

The magnitudes of the three charges are:

q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C

The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:

F_A = F_{BA} + F_{CA} = k\frac{q_B q_A}{(x_B-x_A)^2}+k\frac{q_C q_A}{(x_C-x_A)^2}=\\=(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(10)^2}+(8.99\cdot 10^9) \frac{(0.5\cdot 10^{-6})^2}{(20)^2}=2.8\cdot 10^{-5} N = 28 \mu N

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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Answer:

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Explanation:

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