Answer:
278 K
Explanation:
The distance of the image observed by th girl is 2.77 cm.
The focal length of the lens is calculated as follows;
F = R/2
where;
R = 9 cm/2 = 4.5 cm
F = R/2
F = 4.5/2
F = 2.25 cm
The distance of the image observed by th girl is caculated as follows;
1/f = 1/v + 1/u
1/v = 1/f - 1/u
1/v = 1/2.25 - 1/12
1/v = 0.3611
v = 2.77 cm
Thus, the distance of the image observed by th girl is 2.77 cm.
Learn more about image formed by lens here: brainly.com/question/6722295
Answer:
500 m
Explanation:
The size of a nucleus is about
Instead, the size of the electron cloud extends at the order of
This means that the ratio between the size of the electron cloud and the size of the nucleus is about
So, the electron cloud is about 10,000 bigger than the nucleus of the atom.
Here we want to build a scale-model atom, in which the nucleus is the size of a tennis ball, so a size of approximately
Therefore, since the proportions must be respected, the electron clouds must be 10,000 bigger, so its size in this model would be:
So, the size of the electron cloud would be 500 m.
The cloud of electrons in a scale-model atom would extend about 100 meters.
The nucleus is a central, positively charged region within an atom. It contains protons, which have a positive electric charge, and neutrons, which are electrically neutral. Nuclei make up the majority of an atom's mass and are held together by the strong nuclear force, while electrons orbit the nucleus in electron shells.
In a scale-model atom where the nucleus is the size of a tennis ball, the cloud of electrons would extend about 100 meters. This can be calculated by comparing the size of the nucleus to the actual size of an atom. The diameter of a tennis ball is approximately 6.7 cm, while the diameter of an atom is on the order of 0.1 nm. By scaling up, we find that the electron cloud would extend around 100 meters.
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45°
90°
180°
50°
I'll report you if you don't actually help. I'd like an actual explanation, please.
Hey
So first we need to know what the direction of the force is, using your right hand rule point your right hand in the direction of the velocity. You're saying its the z direction, not telling me whether it's into the page or out? Since its a positive z im assuming its coming out. The magnetic field is pushing it upwards, so the force is going in the negative x direction.
The force of a magnetic field is
F = Qv X B
What's weird is that you don't need mass in this equation. Actually you don't even need the formula, its telling you that they're all going in perpendicar directions. the answer is 90 degrees.
Now if you want to know the F just multiply the charge, velocity and magnetic field .
F = GVB
F = 6.048 E -15
Answer : 90 degrees, sin(90) = 1
To find the magnitude and direction of the magnetic force on a proton moving in a magnetic field, you can use the equation F = qvBsinθ, where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. The magnitude of the magnetic force can be calculated using the equation, and its direction can be determined using the right-hand rule. In this case, the angle between the proton's velocity and the magnetic field is 90°.
To determine the magnitude of the magnetic force on the proton, we need to use the equation F = qvBsinθ, where F is the force, q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.
Plugging in the values, we have F = (1.6 × 10-19 C)(1.8 × 105 m/s)(2.1 × 10-1 T)sinθ.
To find the angle θ, we can use the fact that the force is perpendicular to both the velocity and the magnetic field, which means that sinθ = 1.
Therefore, the magnitude of the magnetic force on the proton is F = (1.6 × 10-19 C)(1.8 × 105 m/s)(2.1 × 10-1 T) = 6.048 × 10-14 N. The direction of the magnetic force is given by the right-hand rule, which shows that the force is perpendicular to both the velocity and the magnetic field, pointing in the positive x-direction.
The angle between the proton's velocity and the magnetic field is 90°.
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