- #1
Kaldanis
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I have my first physics exam coming up in January. To study I've been doing the past exams from the previous years, but there were are a few questions that I wasn't 100% sure about. I'd be greatful if anyone could tell me if my answers are correct or not
Now, I know that if the mass is constant then equation iii) can be used. That makes me think the answer is C. But then again, if the mass is constant I could technically use i) and ii) if I wanted to aswell? So the answer is D.?
Since she's doing half the work she 'should' be doing, the y-component must be taking half the force, or 20N. So is this really as simple as sinθ = 20/40, meaning the answer is A. 30°?
Differentiating with respect to X gives 12xy, and with respect to Y gives 6x2 + 24y3. The only answer with these values is answer E, but E gives negative forces. Is this because gravitational potential energy (U) is negative? Or is my answer just wrong?
1) Which of the following forms of Newton’s Second Law may be used when considering a body of constant mass?
i) [itex]F=m\vec{a}[/itex]
ii) [itex]F=m\frac{d\vec{v}}{dt}[/itex]
iii) [itex]F=\frac{d\vec{p}}{dt}[/itex]
A. (i)
B. (ii)
C. (iii)
D. (i), (ii) and (iii)
E. None apply
i) [itex]F=m\vec{a}[/itex]
ii) [itex]F=m\frac{d\vec{v}}{dt}[/itex]
iii) [itex]F=\frac{d\vec{p}}{dt}[/itex]
A. (i)
B. (ii)
C. (iii)
D. (i), (ii) and (iii)
E. None apply
Now, I know that if the mass is constant then equation iii) can be used. That makes me think the answer is C. But then again, if the mass is constant I could technically use i) and ii) if I wanted to aswell? So the answer is D.?
2) A woman pulls a car 50 m along the ground using a tow-rope attached to
the front of the car. If she does 1000 J of work, exerting a force of 40 N, at
what angle is the rope to the ground?
A. 30°.
B. 45°.
C. 60°.
D. 75°.
E. 90°.
the front of the car. If she does 1000 J of work, exerting a force of 40 N, at
what angle is the rope to the ground?
A. 30°.
B. 45°.
C. 60°.
D. 75°.
E. 90°.
Since she's doing half the work she 'should' be doing, the y-component must be taking half the force, or 20N. So is this really as simple as sinθ = 20/40, meaning the answer is A. 30°?
3) The potential energy for a two-dimensional force is given by the formula
U= 6x2 y + 6y4 . What is the force acting at the point (x, y)?
A. F = (12x)i + (6y3 )j N.
B. F = (6xy + 6y4 )i + (6x2 + 18y3 )j N.
C. F = 6x3 y2 + 6xy5 N.
D. F = 12xy + 24y3 N.
E. F = (-12xy)i + (-6x2 - 24y3 )j N.
U= 6x2 y + 6y4 . What is the force acting at the point (x, y)?
A. F = (12x)i + (6y3 )j N.
B. F = (6xy + 6y4 )i + (6x2 + 18y3 )j N.
C. F = 6x3 y2 + 6xy5 N.
D. F = 12xy + 24y3 N.
E. F = (-12xy)i + (-6x2 - 24y3 )j N.
Differentiating with respect to X gives 12xy, and with respect to Y gives 6x2 + 24y3. The only answer with these values is answer E, but E gives negative forces. Is this because gravitational potential energy (U) is negative? Or is my answer just wrong?