A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?
1 answer:
It will take 6.42 s for the ball that is dropped from a height of 206 m to reach the ground.
From the question given above, the following data were obtained:
Height (H) = 206 m
<h3>Time (t) =? </h3>NOTE: Acceleration due to gravity (g) = 10 m/s²
The time taken for the ball to get to the ground can be obtained as follow:
H = ½gt²
206 = ½ × 10 × t²
206 = 5 × t²
Divide both side by 5
Take the square root of both side
Therefore, it will take 6.42 s for the ball to get to the ground.
Learn more: brainly.com/question/24903556
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88 m/s
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u is the initial velocity
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d = 484 m is the stopping distance
v = 0 is the final velocity
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Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
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x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
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=
- gt
= 0 - g √2y/g
= - √2gy
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θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
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Answer:
The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.
at r < R:
Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.
E = 0.
at R < r < 2R:
The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.
at 2R < r:
The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.
Explanation:
Gauss’ Law is straightforward when applied to spheres. The area of the sphere is , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.
