Answer:
2 x 10⁻³ volts
Explanation:
B = magnetic of magnetic field parallel to the axis of loop = 1 T
= rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = Angle of the magnetic field with the area vector = 0
E = emf induced in the loop
Induced emf is given as
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
(1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
Answer:
The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.
Explanation:
Kepler's three laws are:
1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii
2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals
3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period
There 3 laws help explaining the following statements:
- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.
-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).
- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.
Instead, the following statement:
<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>
Is not supported by any Kepler's law.
Answer:
If both the radius and frequency are doubled, then the tension is increased 8 times.
Explanation:
The radial acceleration (), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:
(1)
Where:
- Frequency, measured in hertz.
- Radius of rotation, measured in meters.
From Second Newton's Law, the centripetal acceleration is due to the existence of tension (), measured in newtons, through the string, then we derive the following model:
(2)
Where is the mass of the object, measured in kilograms.
By applying (1) in (2), we have the following formula:
(3)
From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:
(4)
If both the radius and frequency are doubled, then the tension is increased 8 times.