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3 years ago

In concave mirror, the size of image depends upon

Physics

2 answers:

irina1246 [14]3 years ago

8 0

Answer:

The distance of the object placed on the principal axis from the concave mirror.

Explanation:

In a concave mirror, the nature of the image formed formed by the object placed in front of the mirror depends on the position of the object placed in from of the mirror. It all depends on the distance between the mirror and the object placed on the principal axis.

The closer the object is to the lens, the more larger or magnified the image formed will be. For example an object placed between the focal point and the pole of a concave produces a much larger image than an object placed beyond the centre of curvature of such mirror.

Dmitrij [34]3 years ago

4 0

Answer:

The location of the object relative to the vertex, V, the focus, F center of curvature, C and infinity

Explanation:

The location of the object relative to the vertex, V, the focus, F center of curvature, C and infinity such that;

The image of an object placed at V, touching the mirror is the upright and the same size as the object and virtual

The image of an object placed between V and F is upright and magnified also virtual

The image of an object placed at the focus is at infinity

The image of an object placed between C and F is real, magnified and inverted

The image of an object placed at C is real, same size and inverted

The image of an object placed between C and infinity is real, diminished and inverted

The image of an object placed at infinity is size zero and real.

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An infrared (IR) and ultraviolet (UV) wave propagating through a vacuum must have the same____

Zinaida [17]

Answer:

D. Wavelength

Explanation:

An infrared (IR) and ultraviolet (UV) wave propagating through a vacuum must have the same wavelength.

3 0

2 years ago

A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont

ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

$s = ut + \frac{1}{2}at^{2}$

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

$a = \frac{Fnet}{m}$

where m = mass of the crate, 20 kg

now, $a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}$

from, $s = ut + \frac{1}{2}at^{2}$

$15 = 0*t + \frac{1}{2}* 3.291 * t^{2}$

15 = 0 + 1.645 $t^{2}$

15 = 1.645 $t^{2}$

$t = \sqrt{\frac{15}{1.645} }$

$t = 3.019$

t = 3.01s (to 3 sig fig)

7 0

3 years ago

Ali mixes 20g of Sodium Chloride in 100g of water at 15 Degree °C. Calculate the mass of the solution

Lera25 [3.4K]

Answer:

The mass of the solution is 120 g.

Explanation:

The mass of the solution is given by:

$m_{sol} = m_{1} + m_{2}$

Where:

$m_{sol}$ : is the mass of the solution

$m_{1}$ : is the mass of the solvent

$m_{2}$ : is the mass of the solute

In the solution, the solvent is the majority compound (in mass) and the solute is the minority (in mass), so the solvent is the water and the solute is sodium chloride.

Hence, the mass of the solution is:

$m_{sol} = m_{1} + m_{2} = 100 g + 20 g = 120 g$