Answer:the maximum Hall voltage across the strip= 0.00168 V.
Explanation:
The Hall Voltage is calculated using
Vh= B x v x w
Where
B is the magnitude of the magnetic field, 5.6 T
v is the speed/ velocity of the strip, = 25 cm/s to m/s becomes 25/100=0.25m/s
and w is the width of the strip= 1.2 mm to meters becomes 1.2 mm /1000= 0.0012m
Solving
Vh= 5.6T x 0.25m/s x 0.0012m
=0.00168T.m²/s
=0.00168Wb/s
=0.00168V
Therefore, the maximum Hall voltage across the strip=0.00168V
This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.
Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.
I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.
The missing diagram is in the attachments.
Answer: X: positive Y: positive
Explanation: Electric field is a vector quantity, which means it can be represented by a vector arrow: the arrow points in the direction of electric field and its length represents the magnitude at a given location. There are another representation of the electric field called electric field lines, <u>in which the line points away from a positively charged source and towards a negatively charged source</u>. This occurs because it follows a pattern, where the lines points in the direction that a positive test charge would have if it is accelerating on the line.
Analyzing the diagram, it can be observed that the lines are pointing away from both of the charged objects. Therefore, both X and Y are <u>positively charged</u>.
An initial velocity is:
v o = 25 m/s
The vertical component of the initial velocity:
v o y = v o * sin 60° =
= v o * √3 / 2 = 25 m/s * √3 / 2 = 21.65 m/s
Answer:
The approximate vertical component of the initial velocity is 21.65 m/s.