C). I think! Hope this helps!!
Well first of all, when it comes to orbits of the planets around
the sun, there's no such thing as "orbital paths", in the sense
of definite ("quantized") distances that the planets can occupy
but not in between. That's the case with the electrons in an atom,
but a planet's orbit can be any old distance from the sun at all.
If Mercury, or any planet, were somehow moved to an orbit closer
to the sun, then ...
-- its speed in orbit would be greater,
-- the distance around its orbit would be shorter,
-- its orbital period ("year") would be shorter,
-- the temperature everywhere on its surface would be higher,
-- if it has an atmosphere now, then its atmosphere would become
less dense, and might soon disappear entirely,
-- the intensity of x-rays, charged particles, and other forms of
solar radiation arriving at its surface would be greater.
Answer:
0.255
Explanation:
The following data were obtained from the question:
Force (F) = 57 N
Mass (m) = 22.8 Kg
Coefficient of static friction (µ) =...?
Next, we shall determine the normal reaction (R). This is illustrated below:
Mass (m) = 22.8 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (R) =?
R = mg
R = 22.8 x 9.8
R = 223.44 N
Finally, we can obtain the coefficient of static friction (µ) as follow:
Force (F) = 57 N
Normal reaction (R) = 223.44 N
Coefficient of static friction (µ) =...?
F = µR
57 = µ x 223.44
Divide both side by 223.44
µ = 57/223.44
µ = 0.255
Therefore, the coefficient of static friction (µ) is 0.255.
Answer:
I2 = 3.076 A
Explanation:
In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:
(1)
I: current in the wire
R: radius of the wire
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:
(2)
I1: current of the first ring = 8A
R1: radius of the first ring = 0.078m
I2: current of the second ring = ?
R2: radius of the first second = 0.03m
To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:
The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.
