How long will it take a rock dropped from 2.0 m above the surface of mercury to reach the ground?
1 answer:
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The electric field at arbitrary point outside the sphere is determined as the E = σr³/k.
<h3>Electric field determined from Gauss law</h3>The electric field of the surface is determined from Gauss law as shown below;
E ∫ds = Q/ε
E (4πr²) = Q/ε
E = Q/4πεr² . r
where;
- σ is dipole moment of atom of the metal
- k is dielectric constant
Thus, the electric field at arbitrary point outside the sphere is determined as the E = σr³/k.
The complete question:
A metal sphere of radius R carries a total charge Q. outside the sphere is a dielectric with polarization p(f) k/r^3er. Determine the electric field at arbitrary point outside the sphere.
Learn more about electric field here: brainly.com/question/14372859
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Answer:
She must be launched with a speed of 74.2 m/s.
Explanation:
Hi there!
The equations of the horizontal component of the position vector and the vertical component of the velocity vector are the following:
x = v0 · t · cos θ
vy = v0 · sin θ + g · t
x = horizontal distance traveled at time t.
v0 = initial velocity.
t = time.
θ = launching angle.
vy = vertical component of the velocity vector at time t.
g = acceleration due to gravity (-9.8 m/s²).
To just cross the 520-m gap, the maximum height of the flight must be reached halfway of the gap at 260 m horizontally (see attached figure).
When she is at the maximum height, her vertical velocity is zero. So, when x = 260 m, vy = 0. Using both equations we can solve the system for v0:
x = v0 · t · cos θ
Solving for v0:
v0 = x/ (t · cos θ)
Replacing v0 in the second equation:
vy = v0 · sin θ + g · t
0 = x/(t·cos(56°)) · sin(56°) + g · t
0 = 260 m · tan (56°) / t - 9.8 m/s² · t
9.8 m/s² · t = 260 m · tan (56°) / t
t² = 260 m · tan (56°) / 9.8 m/s²
t = 6.27 s
Now, let's calculate v0:
v0 = x/ (t · cos θ)
v0 = 260 m / (6.27 s · cos(56°))
v0 = 74.2 m/s
She must be launched with a speed of 74.2 m/s.
