Answer:
The magnetic force points in the positive z-direction, which corresponds to the upward direction.
Option 2 is correct, the force points in the upwards direction.
Explanation:
The magnetic force on any charge is given as the cross product of qv and B
F = qv × B
where q = charge on the ball thrown = +q (Since it is positively charged)
v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)
B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)
F = qv × B = (+qvî) × (Bj)
F =
| î j k |
| qv 0 0|
| 0 B 0
F = i(0 - 0) - j(0 - 0) + k(qvB - 0)
F = (qvB)k N
The force is in the z-direction.
We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.
Hope this Helps!!!
Answer:
y = 0.0233 m
Explanation:
In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:
Δx = λL/d
where,
Δx = distance between fringes
λ = wavelength of light
L = Distance between screen and slits
d = Slit Separation
Now, for initial case:
λ = 425 nm = 4.25 x 10⁻⁷ m
y = 2Δx = 0.0177 m => Δx = 8.85 x 10⁻³ m
Therefore,
8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d
L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)
L/d = 2.08 x 10⁴
using this for λ = 560 nm = 5.6 x 10⁻⁷ m:
Δx = (5.6 x 10⁻⁷ m)(2.08 x 10⁴)
Δx = 0.0116 m
and,
y = 2Δx
y = (2)(0.0116 m)
<u>y = 0.0233 m</u>
Answer:
im not too sure about that all i know is history
Answer:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:
Solving for the force F, we obtain that:
Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:
From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:
In this case, the force from the rocket's thrusters is equal to 118,000N.
In order to make his measurements for determining the Earth-Sun distance, Aristarchus waited for the Moon's phase to be exactly half full while the Sun was still visible in the sky. For this reason, he chose the time of a half (quarter) moon.
<h3 /><h3>How did Aristarchus calculate the distance to the Sun?</h3>
It was now possible for another Greek astronomer, Aristarchus, to attempt to determine the Earth's distance from the Sun after learning the distance to the Moon. Aristarchus discovered that the Moon, the Earth, and the Sun formed a right triangle when they were all equally illuminated. Now that he was aware of the distance between the Earth and the Moon, all he needed to know to calculate the Sun's distance was the current angle between the Moon and the Sun. It was a wonderful argument that was weakened by scant evidence. Aristarchus calculated this angle to be 87 degrees using only his eyes, which was not far off from the actual number of 89.83 degrees. But when there are significant distances involved, even slight inaccuracies might suddenly become significant. His outcome was more than a thousand times off.
To know more about how Aristarchus calculate the distance to the Sun, visit:
brainly.com/question/26241069
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