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2 years ago

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A convex mirror with a radius of curvature of 0.550m is placed above the aisles in a store. Determine the image distance and mag

nification of a customer lying on the floor 3.1 m below the mirror. Is the image virtual or real? Is the image inverted or upright?

1 answer:

Otrada [13]2 years ago

8 0

Answer:

Explanation:

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As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. incre

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

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Calculate the mass of air in a room of floor dimensions =10M×12M and height 4m(Density of air =1.26kg/m cubic

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The volume of the room is the product of its dimensions:

$10\times 12 \times 4 = 480\text{ m}^3$

Now, from the equation

$d=\dfrac{m}{V}$

where d is the density, m is the mass and V is the volume, we deduce

$m=dV$

So, multiply the density and the volume to get the mass of air in the room.

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Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark

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Answer:

Wavelength of incident light, $\lambda = 612 nm$

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

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Solution:

To calculate the wavelength of the incident light, $\lambda$ :

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

$\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm$

$\lambda = 612 nm$

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