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melomori [17]
3 years ago
11

Vector a with arrow has a magnitude of 12.2 units and points due west. vector b with arrow points due north. (a) what is the mag

nitude of b with arrow if a with arrow + b with arrow has a magnitude of 17.4 units? 21.2 incorrect: your answer is incorrect. units (b) what is the direction of a with arrow + b with arrow relative to due west? ° correct: your answer is correct. (c) what is the magnitude of b with arrow if a with arrow â b with arrow has a magnitude of 17.4 units? units (d) what is the direction of a with arrow â b with arrow relative to due west? ° correct: your answer is correct.
Physics
1 answer:
Molodets [167]3 years ago
8 0
C = Sqrt (17.4^2 - 12.2^2) = 12.4
ArcTan (12.4/12.2) = 45.47 degrees above vector a,  which is 134.53 degrees measured on the xy-axis.

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Help! PROJECTILE PROBLEM: A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25
dsp73
Here, 

height at failure, h1 = 525 m, 
upward acceleration, a = 2.25 m/s^2, 
velocity = v m/s, 
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SO, </span>
<span>
v^2 = 2*a*h = 2*2.25*525 = 2362.5 </span>
Now, acceleration, g = 9.8 m/s^2, 
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heigt, h1 = v^2/2g = 2362.5 / 2*9.8 = 120.54 meters </span>
Hence, 
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a) </span>
Total height = 525+120.54 = 645.54 meters 

b) 
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4 0
3 years ago
Read 2 more answers
Use the equation for magnetic force on a moving charge to derive the equation for magnetic force on a current carrying wire. Sho
max2010maxim [7]

Answer:

The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F  is force and is a vector because the force acts in a direction.  q  is the charge of the particle.  v  is velocity and is a vector because the particle is moving in some direction.  B is the magnetic flux density.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd.  Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore

                                                  ∮B dl=μ0 I

                                                  B ∮dl=μ0 I

                                                  B 2πr=μ0 I

                                                   B=μ02πr Id=μ0/4π I dl×rr3

Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get

                                               dl×r=−z dlj^+rsinθk^

and therefore the magnitude of dB is equal to

dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3

Thus, magnetic field is depending on r,θ,z.

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7 0
2 years ago
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. if the c
goldfiish [28.3K]
The bearing could be the below: 
oppositely charged, same initial direction 
same charge, opposite initial direction

You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.
4 0
4 years ago
A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit
Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E

now from above equation

P = 4390 N/m^2

\rho = 0.999 * 10^3

v = 7.22 * 10^{-4} m/s

H = 23.6 m

now by above equation

\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E

E = 235.7 J/kg

Part B)

Now energy per unit weight

U = \frac{E}{g}

U = \frac{235.7}{9.8}

U = 24 m

7 0
4 years ago
A bike accelerates uniformly(from rest to a speed of 7 m/s over a distance of 40 m. Determine
marshall27 [118]

Answer:

0.61 m/s^2

Explanation:

The bike's acceleration can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity of the bike

u is the initial velocity

a is the acceleration

s is the distance covered

For the bike in the problem,

u = 0

v = 7 m/s

d = 40 m

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{7^2-0}{2(40)}=0.61 m/s^2

3 0
3 years ago
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