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ivanzaharov [21]

3 years ago

6

what type of image is formed by a lens if m = -4.6? A. One that is larger than the object and virtual, B. One that is smaller th

an the object and virtual, C. One that is larger then the object and real, D. One that is smaller than the object and real

1 answer:

Montano1993 [528]3 years ago

8 0

Answer:

One that is larger than the object and real

Explanation:

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As a pendulum swings from its highest to lowest position, what happens to its kinetic and potential energy?

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The potential energy decreases while the kinetic energy increases.

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Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

4 years ago

A cylinder contains 3.5 L of oxygen at 350 K and 2.7 atm . The gas is heated, causing a piston in the cylinder to move outward.

Effectus [21]

Answer:

The pressure is $P_2 = 1.84 \ a.t.m$

Explanation:

From the question we are told that

The first volume of is $v_1 = 3.5 \ L$

The first pressure is $P_1 = 2.7 \ a.t.m$

The first temperature is $T_1 = 350 \ K$

The new temperature is $T_2 = 620 \ K$

The new volume is $V_2 = 9.1 \ a.t.m$

Generally according to the combined gas law we have that

$\frac{P_1 V_1 }{T_1 } = \frac{P_2 V_2 }{T_2 }$

=> $P_2 = \frac{P_1 * V_1 * T_2 }{T_1 * V_2 }$

=> $P_2 = \frac{ 2.7 * 3.5 * 620 }{ 350 * 9.1 }$

=> $P_2 = 1.84 \ a.t.m$

4 0

4 years ago