I thought the convention would be vector A- vector B= vector A + (-vector B), or 3.00units + (-4.0)units = -1.0 units.

but this isn't the same as the answer provided in thesolutions manual. I also don't understand how the graph got flippedif vector A is along the positive x-axis and vector B is along thenegative Y axis.

thanks.

I also don't understand how to know which angle I am measuringin order to find the direction of vectoraddition/subtraction.

answer

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Nestor RutherfordLv2

8 Mar 2019

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Someone posted a response to this earlier but I don't think theanswer was correct... Can someone please take the time to help mecarefully with this? Thanks!

A kayaker can paddle her kayak at a steady 2.5m/s in still water.She wishes to cross a river that is 2000m wide and has a current of-1.5m/s (goes from right to left along x axis). She is crossinggoing "north", along the positive y axis.

(a) IF the kayaker first "aims" her craft straight across theriver, the current will carry her downstream as she paddles across.What will be her actual velocity (magnitude and direction angle) asshe crosses? How long will it take her to cross the river?

(b) If she "aims" the kayak somewhat upstream, she can actuallytravel straight accross the river. In what direction must she aim?What is her actual speed accross the river for this situation, andhow long will it take her to cross?

HINT: The angle found in (a) will NOT be the same angle required inpart (b).

--- I have figured out (a) to be: a velocity of 2.92 m/s at121 degrees (CCW from x axis). I figured it would take 805s tocross (the distance she travels, or the hypotonuse of the triangle,is 2333.27m).

I really need help on part (b) though :(

Is that right? And could someone help me on (b) please? :)

A mass m rests on a frictionless horizontal table and isconnected to rigid supports via 2 identical springs each of relaxedlength L0 and spring constantk. Each spring is stretched to a length L considerable greater thanL0. Horizontal displacementsof m from its equilibrium position are labeled x and y.

(Diagram insert didn't work, all it really tells you is that xis on the axis of the springs and y is perpendicular to thesprings)

a) Write down the differential equation of motion (i.e.,Newton's Law) governing small oscillations in the xdirection.

b) Same as a but in the y direction (assume that y <<L).

c) In terms of L and L0, calculate the ratio of the periods ofthe oscillations along x and y.

d) If at t=0 the mass m is released from the point x = y = A0with zero velocity, what are its x and y coordinates at any latertime t?

---------------------------

So, I started with just ma = -2kx for a, leading me to thedifferential equation x''(t) + (2k/m) x = 0.

Then I did basically the same thing with y.

However, when I get to c, there's my problem - I don't knowhow I could write any of this in L and L0.

Also, b doesn't really make sense to me because if y is smallwhen compared to L, it seems like there shouldn't be anyoscillation in the y direction.

The answer the back of the book shows for c and d are:

c) Tx/Ty = (1 - L0 / L)1/2

d) x(t) = A0cos( (2k/m)1/2 t)

y(t) = A0cos( (2k(L-L0)/mL)1/2 t)

I have no idea how they got to these answers. Can somebodyhelp me? It has something to do with springs already beingstretched, but it seems they are making an assumption somewherethat I don't understand.

Goal Use the superposition principle to calculate the electricfield due to two point charges.

Problem Charge q1 = 7.00 ÂµC is at the origin, and charge q2 = -5.35ÂµC is on the x-axis, 0.300 m from the origin (Fig. 15.12).

(a) Find the magnitude and direction of the electric field at pointP, which has coordinates (0, 0.400) m.

(b) Find the force on a charge of 2.00 10-8 C placed at P.

(a) Calculate the electric field at P.
Find the magnitude of 1 with Equation 15.6.
E1 = 3.93 105 N/C
Vector 1 is vertical, making an angle of 90Â° with respect to thepositive x-axis. Use this fact to find its components. E1x = E1cos(90Â°) = 0
E1y = E1 sin(90Â°) = 3.93 105 N/C
Next, find the magnitude of 2, again with Equation 15.6.
E2 = 1.92 105 N/C
Obtain the x-component of 2, using the triangle in Figure 15.12 tofind cos ?.
E2x = E2 cos ? = (1.92 105 N/C)(0.600)
E2x = 1.15 105 N/C
Obtain the y-component in the same way, but a minus sign has to beprovided for sin ? because this component is directeddownward.
E2y = E2 sin ? = (1.92 105 N/C)(0.800)
E2y = -1.54 105 N/C
Sum the x-components to get the x-component of the resultantvector. Ex = E1x2x = 0 + 1.15 105 N/C
Ex = 1 N/C
+ E
Sum the y-components to get the y-component of the resultantvector. Ey = E1y2y = 3.93 105 N/C - 1.54 105 N/C
Ey = 2 N/C
+ E
Use the Pythagorean theorem to find the magnitude of the resultantvector.
E = 3 N/C

The inverse tangent function yields the direction of the resultantvector.
? = 4Â°

(b) Find the force on a charge of 2.00 10-8 C placed at P.
Calculate the magnitude of the force (the direction is the same asthat of because the charge is positive). F = Eq = E(2.00 10-8C)
F = 5 N

(a) Place a charge of -3.60 ÂµC at point P and find the magnitudeand direction of the electric field at the location of q2.

E = N/C

? = Â° counterclockwise from the positive x-axis

(b) Find the magnitude and direction of the force on q2.

F = N

? = Â° counterclockwise from the positive x-axis

I Don't know how to do this Please help Me. Thank you. The top partis how i figured out the rest i think its the missing informationto solve the bottom part.

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