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3 years ago

A ray diagram without the produced image is shown.

1 answer:

8 0

A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.

 The image produced by the lens is (b) inverted and real

Explanation:

A real image occurs where the rays converge.

Real images can be produced both by the concave mirrors or converging lenses, but the condition is that the object of consideration is always placed far away from the mirror or the lens than the focal point, and thus the real image produced is inverted.

 The image produced by the lens is (b) inverted and real

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In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:

vesna_86 [32]

Answer:

D. 1×10⁻⁸ M

Explanation:

[H⁺] [OH⁻] = 10⁻¹⁴

(1×10⁻⁶) [OH⁻] = 10⁻¹⁴

[OH⁻] = 1×10⁻⁸

6 0

3 years ago

What is the momentum of a 12 kg condor flying at 6 m/s?

rusak2 [61]

Answer:

Explanation:

Formula

$p=mv\\=12*6\\=72$

5 0

3 years ago

How much force does an 88kg astronaut exert on his chair while accelerating straight up at 10 m/s^2

Fed [463]

Force=mass*acceleration. So 88kg*10 m/s^2=880 newtons

6 0

3 years ago

wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o

tresset_1 [31]

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

(b) What is the total charge stored in the parallel combination?

Answer:

i $Q_1 = 2.124 *10^{-11} \ C$

ii $Q_2 = 4.4604 *10^{-11} \ C$

$Q_{eq} = 6.5844 *10^{-11} \ C$

Explanation:

From the question we are told that

The voltage of the battery is $V = 12.0 \ V$

The plate area of each capacitor is $A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2$

The separation between the plates is $d = 2.65 \ mm = 2.65 *10^{-3} \ m$

The permittivity of free space has a value $\epsilon_o = 8.85 *10^{-12} \ F/m$

The dielectric constant of the other material is $z = 2.10$

The capacitance of the first capacitor is mathematically represented as

$C_1 = \frac{\epsilon * A }{d }$

substituting values

$C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 1.77 *10^{-12} \ F$

The charge stored in the first capacitor is

$Q_1 = C_1 * V$

substituting values

$Q_1 = 1.77 *10^{-12} * 12$

$Q_1 = 2.124 *10^{-11} \ C$

The capacitance of the second capacitor is mathematically represented as

$C_2 = \frac{ z * \epsilon * A }{d }$

substituting values

$C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 3.717 *10^{-12} \ F$

The charge stored in the second capacitor is

$Q_2 = C_2 * V$

substituting values

$Q_2 = 3.717*10^{-12} * 12$

$Q_2 = 4.4604 *10^{-11} \ C$

Now the total charge stored in the parallel combination is mathematically represented as

$Q_{eq} = Q_1 + Q_2$

substituting values

$Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}$

$Q_{eq} = 6.5844 *10^{-11} \ C$

7 0

3 years ago

After falling from rest at a height of 28.7 m, a 0.502 kg ball rebounds upward, reaching a height of 19.8 m. If the contact betw

Ad libitum [116K]

Answer:

9080 N

Explanation:

Consider the two motions of the ball.

In the downward motion, initial velocity, u, is 0 (because it falls from rest) and the distance is 28.7 m. Using the equation of motion and using g as 9.8 m/s²,

v² = u² + 2as

v² = 0² + 2 × 9.8 × 28.7 = 562.52

v = 19.7 m/s

For the downward motion, the initial velocity is unknown, the final velocity is 0 and initial velocity is desired. g is negative because the motion is upwars.

0² = v² - 2 × 9.8 × 19.8

v² = 388.08

v = 10.7 m/s

The change in momentum = 0.502(10.7 -(23.7)) = 21.7868 kgm/s

The impulse = change in monetum

Ft = 21.7868 kgm/s

But t = 2.4 ms

$[tex]F = \dfrac{21.7868}{2.4\times10{-3}} = 9078 \text{ N}$ [\tex]

8 0

3 years ago