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Answer:
r= 2.17 m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 2.9 N/C
λ = 3.5*10⁻¹⁰ C/m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2*k*(λ/r)
2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)
r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)
r= 2.17 m
<span>1.5 minutes per rotation.
The formula for centripetal force is
A = v^2/r
where
A = acceleration
v = velocity
r = radius
So let's substitute the known values and solve for v. So
F = v^2/r
0.98 m/s^2 = v^2/200 m
196 m^2/s^2 = v^2
14 m/s = v
So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin.
Its circumference is 2*pi*r, so
C = 2 * 3.14159 * 200 m
C = 1256.636 m
And we need a velocity of 14 m/s, so
1256.636 m / 14 m/s = 89.75971429 s
Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
Explanation:
A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path subjected to gravity force only.
Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.
In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.
Answer:
t = 103.45 n m
Explanation:
given,
refractive index of cornea = 1.38
refractive index of eye drop = 1.45
wavelength of refractive index = 600 nm
refractive index of eye drop is greater than refractive index of cornea and the air.
Formula used in this case
for constructive interference

At m = 0 for the minimum thickness, so
t = 103.45 n m
the minimum thickness of the film of eyedrops t = 103.45 n m