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Setler79 [48]
3 years ago
7

If a negatively charged particle moves into a magnetic field traveling in a straight line, how would you expect its motion to ch

ange? A) The particle would slow and stop moving. B) The particle would travel in a curved path. C) The particle would continue traveling in a straight line. D) The particle would make a right angle turn and then continue traveling in a straight line.
Physics
2 answers:
gayaneshka [121]3 years ago
7 0
Hello
the answer is a
Yanka [14]3 years ago
3 0

Answer:

B) The particle would travel in a curved path.

Explanation:

I just took the test and this is the correct answer

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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

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Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in
sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, P_{atm} = 101 Kpa

P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}

P_{gauge}  = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

P_{gauge}  = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge}  = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge}  = 110,298.26 pa

Therefore, the gauge pressure of air is 110 kpa

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