Answer:
2.40 x 10⁻¹³ C
Explanation:
= number of electrons = 6.25 x 10⁶
= charge on electron = - 1.6 x 10⁻¹⁹ C
= number of protons = 7.75 x 10⁶
= charge on proton = 1.6 x 10⁻¹⁹ C
Net charge is given as
Q = +
Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)
Q = 2.40 x 10⁻¹³ C
Answer:
w = -101 rad / s
Explanation:
For this exercise we will use Faraday's law
E = - dФ / dt
where the magnetic flux is
Ф = B. A = B A cos θ
In this case, the angle between the magnetic field and the normal to the disk is zero, cos 0 = 1, they indicate that the field is constant, let's find the area
The area rotated by the disk is
A = ½ r s
if we express the angles in radians
θ = s / r
s = r θ
where is the arc supported
A = ½ r (r θ)
let us substitute in the Faraday equation
E = - d (B ½ r² θ) / dt
E = - ½ B r² dθ/dt
the definition of angular velocity is
w = dθ/dt
E = - ½ B r² w
w = - 2E / B r²
let's calculate
w = - 2 3.86 / (0.0314 1.56²)
w = -101 rad / s
Answer:
Explanation:
Angular speed of hoop ω = v / r
= 8.90 / .27
= 32.96 rad / s
Translational kinetic energy = 1/2 mv²
= .5 x 1.8 x 8.9²
= 71.29 J
Rotational kinetic energy = 1/2 Iω²
= 1/2 mR²x ω²
= 1/2 mv²
= 71.29 J
Total kinetic energy
= 2 x 71.29
= 142.58 J
This energy will be used to attain height
If h be the height attained
mgh = 142.58
h = 142.58 / mg
= 142.58 / 1.8 x 9.8
= 8.08 m .
Answer:
112 km
Explanation:
brainliest plzzzzz i've never got one
Answer: 0.313 rad/s
Explanation:
The equation that relates the velocity and the angular velocity in the uniform circular motion is:
(1)
Where is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.
Isolating from (1):
(2)
On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration is , and is given by the following equation:
(3)
Isolating :
(4)
(5)
Substitutinng (5) in (2):
(6)
This is the angular velocity that would produce an “artificial gravity” of 9 .