. An electron relocating at 4.00×103m/s in a 1.25-T magnetic ar experiences a magnetic pressure of 1.40×10−16N . What edge does the


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hunterriverpei.com: 9.59° and also 350.41°

Explanation: The formulae that relates the pressure F exerted ~ above a moving charge q with velocity v in a magnetic field of toughness B is offered as

F =qvB sin x

Where x is the angle in between the stamin of magnetic field and velocity that the charge.

q = 1.609×10^-19 C

v = 4×10³ m/s

B = 1.25 T

F = 1.40×10^-16 N

By substituting the parameters, we have that

1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx

sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25

sin x = 1.40×10^-16 /8.045*10^(-16)

sin x = 0.1666

x = 9.59°

The worth of sin x is confident in very first and 4th quadrant.

Hence to obtain the second value of x, we move to the fourth quadrant of the trigonometric quadrant i beg your pardon is 360 - x

Hence = 360 - 9.59 = 350.41°


5 0
1 year ago

12) go the brand “centripetal force” ever show up in an FBD?
Murrr4er <49>
\"Though a force, or a resulting sum of forces, may be centripetal, over there is no an really \"Centripetal Force\" that would show up on an FBD. Because an FBD is a diagram of physical pressures at work upon a body, and also a centripetal force is not the reaction of part physical object upon another, it would not be suitable to display it. You can have a tension force, a load force, a normal force, electric forces, magnetic forces, or any number of reactions, however actually labeling any kind of of them \"centripetal\" would be unconventional\".-hunterriverpei.comhoo
6 0
11 months ago

If a glass is knocked turn off of a table that is 1.4m tall, just how long does the take because that the glass to hit the ground?
Lapatulllka <165>
H = 1.4 m, the elevation the glass falls.u = 0, initial upright velocity, since the glass is knocked off a table.Assume g = 9.8 m/s², and also ignore wait resistance.Let t = the moment for the glass to hit the ground.Then(1/2)*(9.8 m/s²)*(t s)² = (1.4 m)4.9t² = 1.4t² = 0.2857t = 0.5345 shunterriverpei.com: 0.54 s (nearest hundredth)
3 0
11 month ago

Consider a particle with early velocity v⃗ that has actually magnitude 12.0 m/s and is directed 60.0 degrees above the an unfavorable x axis.
lina2011 <118>

hunterriverpei.com:

-6.0 m/s, 10.4 m/s

Explanation:

To discover the x- and also y- components, we have actually to apply the formulas:

\"*\"

\"*\"

where

v = 12.0 m/s is the size of the vector

\"*\"
is the angle in between the direction that the vector and also the optimistic x-axis

Here, the angle given is the angle above the an adverse x-axis; this method that the angle with respect come the confident x-axis is

\"*\"

So, the two materials are:

\"*\"

\"*\"


5 0
9 months ago

A planet has two moons. The very first moon has an orbital period of 1.262 planet days and an orbital radius of 2.346 x 104 km. The se
Maslowich
Kepler\"s third law hypothesizes that for all the tiny bodies in orbit about the same main body, the ratio of (orbital period squared) / (orbital radius cubed) is the same number. Moon #1: (1.262 days)² / (2.346 x 10^4 km)³Moon #2: (orbital period)² / (9.378 x 10^3 km)³Equating the ratio:(1.262 days)² / (2.346 x 10^4 km)³ = (orbital period)² / (9.378 x 10^3 km)³Cross-multiply:(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³Divide every side by (2.346 x 10^4)³:(Orbital period)² = < (1.262 days)² x (9.378 x 10^3)³ > / (2.346 x 10^4)³ = 0.1017 day²Orbital period = 0.319 earth day = about 7.6 hours.
4 0
1 year ago
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