Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?
a. it is reduced by a factor of 3
The answer is the less dense plate slides over the denser plate.
Answer:
Explanation:
Project mass m=3.8 kg
Initial speed vi= 0m/s
Final speed vf= 9.3×10³ m/s
Force F=9.3×10⁵N
To find
Time t
Solution
From Newtons second law we know that
∑F=ma
Where m is mass
a is acceleration
We can write this equation as
∑F=m(Δv/Δt)
Rearrange this equation to find time t
So
Substitute the given values
The speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
<h3>
Speed of the satellite</h3>
v = √(GM/r)
where;
- G is universal gravitation constant
- M is mass of Earth
- r is radius of the satellite
v = √(6.67 x 10⁻¹¹ x 5.98 x 10²⁴/3.57 x 6.37x 10³)
v = 1.32 x 10⁵ m/s
Thus, the speed of the satellite in a circular orbit around the Earth is 1.32 x 10⁵ m/s.
Learn more about speed of satellite here: brainly.com/question/22247460
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