## GR0877 Solutions cloud

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## GR0877 Problem 1

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) Once released, the ball experiences only the gravitational force (in this problem we ignore friction). Since the gravitational force has no horizontal component, the ball is moving in a straight line when viewed from above. Thus one should eliminate all but choices (B) and (D). The car is moving to the right, while the ball is thrown perpendicularly to this direction, so the initial velocity of the ball is directed towards southeast and preserves this direction in the future due to the absence of horizontal forces able to change this direction (Newton’s 2nd law). Choice (B).

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Categories: PGRE Tags: , , , ,

## GR0877 Problem 2

10.10.2011 1 comment

PROBLEM STATEMENT: This problem is still being typed .

SOLUTION: (D) If we ignore friction, horizontal and vertical motions of the object are independent. Therefore, no one really needs the initial horizontal component of the velocity to determine how much it takes to cover some distance in the vertical direction. The initial vertical component of the velocity is zero, so from $H = gt^2/2$ one gets $H = (9.8 \cdot 2^2 )/2 = 9.8 \cdot 2 = 19.6$ (m).

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The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: PGRE Tags: , , , ,

## GR0877 Problem 3

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) The dissipating power is $P = V^2/R$, where $V$ is the voltage across resistor $R$. Hence, if you double $U$, the power will quadruple. Comment: you cannot use $P = VI$, because if you change $V$ the current $I$ through the resistor will also change.

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The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: PGRE Tags: , , , ,

## GR0877 Problem 4

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) The loop lies exactly on a magnetic field line of the long wire. Thus, at every point of the loop the magnetic field due to the wire is parallel to the direction of the current $I_2.$ Thus, the magnetic force on the loop is $\mathbf{F}_{mag} = I_2 \int (d\mathbf{l} \times \mathbf{B}) = 0$, because the cross product of two parallel vectors is zero.

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Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: PGRE Tags: , , , ,

## GR0877 Problem 5

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (A) The de Broglie relation is $\lambda = h/p$, where $h$ is Planck’s constant.

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Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

Categories: PGRE Tags: , , , ,

## GR0877 Problem 6

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) The $n$-th shell can accommodate $2n^2$ electrons. The factor 2 is due to the fact that each orbital can have at most two electrons (one spin up and one spin down). Thus, the total number of electrons is $2(1^2 + 2^2) = 2 \cdot 5 = 10$.

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