Simple Vector Problem - Finding the x-component

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To find the x-component of a velocity vector at 45 degrees below the positive x-axis with a y-component of -29, the correct approach involves using trigonometric functions. The initial calculations were incorrect due to a misunderstanding of the sine function application. The x-component can be derived from the equation Vx = |V|Cos(Theta) after properly determining the magnitude of the vector. For a vector moving in the negative x-direction at 7.0 cm/s, the x-component is -7.0 cm/s, and the y-component is 0, confirming the direction of the vector. Clarifications on angle usage and direction are essential for accurate component calculations.
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Homework Statement



A velocity vector 45 below the positive x-axis has a y-component of - 29 .
What is the value of its x-component?

Homework Equations



Vx = |V|Cos(Theta)
Vy = |v|Sin(Theta)

The Attempt at a Solution



Theta = -45 degrees
y-component: -29 = xSin(-45) ... where x is the magnitude of the vector

x = Sin(-45) / - 29
...x = 0.2438

So |V| = 0.24~

Vx = (0.24)Cos(-45)
...=0.171

Says it's wrong, and I don't know where to go now. Am a bit embarassed asking for help with a simple problem like this :\
 
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Oops sorry, nevermind, I realized that in Step 2 I should have divided by Sin(-45)

apologies!
 

Homework Statement



Find x- and y-components of the following vectors
v = 7.0 cm/s, negative x-direction

Homework Equations



Vx = |V|Cos(Theta)
Vy = |v|Sin(Theta)

The Attempt at a Solution



I'm a bit stuck here. I slotted in 7cm/s for |v| but how do I find the components if I'm not given an angle or graphical representation?
 
It says 'negative x direction'. I think that means it points along the x-axis in the, uh, 'negative' direction. Are you sure you don't want to rethink posting this question?
 
I thought of that after I posted it and came up with -1, 0 as the x,y components [using cos180 and sign 180) which was still wrong. it's late so I wasn't thinking clearly :P
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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