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4 years ago

What condition is necessary for an object to make a good reference point?

Physics

1 answer:

777dan777 [17]4 years ago

8 0

The object is fixed relative to the motion you are trying to describe.

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Monochromatic light with a wavelength of 384 nm passes through a single slit and falls on a screen 86 cm away. If the distance o

valkas [14]

This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
$y= \frac{m\lambda D}{a}$
from the center of the pattern. In the formula, m is the order of the minimum, $\lambda$ the wavelenght, $D$ the distance of the screen from the slit and $a$ the width of the slit.

In our problem, the distance of the first-order band (m=1) is $y=0.22 cm$ . The distance of the screen is D=86 cm while the wavelength is $\lambda = 384 nm=384 \cdot 10^{-7}cm$ . Using these data and re-arranging the formula, we can find a, the width of the slit:
$a= \frac{m \lambda D}{y}= \frac{1 \cdot 384 \cdot 10^{-7}cm \cdot 86 cm}{0.22 cm}=0.015 cm$

3 0

3 years ago

Write in the word "stronger" or "weaker." The bigger and

SVETLANKA909090 [29]

Answer:

weaker has the heavier of an object

4 0

3 years ago

For what absolute value of the phase angle does a source deliver 71 % of the maximum possible power to an RLC circuit

kherson [118]

Answer: the absolute value of the phase angle is 28°

Explanation:

taking a look at expression for the instantaneous electric power in an AC circuit;

P = VI -------let this be equation 1

p is power, v is voltage and I is current;

for maximum power

P_max = V_rms × I_rms --------let this be equ 2

where P_max is the maximum power, V_rms is the rms value of voltage and I_rms is the rms value of current.

Also for average electric power in an AC circuit

P_avg = V_rms × I_rms × cos²∅ -------let this be equ 3

where P_avg is the average power and cos∅ is the power factor

now from equation 2; P_max = V_rms × I_rms

so p_max replaces V_rms × I_rms in equation 3

we now have

P_avg = P_max × cos²∅

so we substitute

expression for the given value of the average power is

P_avg = P_max × 75%

p_avg = P_max.78/100

for the expression of the average electricity in an AC circuit

P_max.78/100 = P_max × cos²∅

78/100 = cos²∅

to get the absolute value of phase angle

∅ = cos⁻¹ ( √(78/100))

∅ = cos⁻¹ ( 0.8832)

∅ = 27.969 ≈ 28°

Therefore the absolute value of the phase angle is 28°

8 0

3 years ago

6. Two light bulbs are designed for use at 120 V and are rated at 75 W and 150 W. Which light bulb has the greater filament resi

Nonamiya [84]

$\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}$

P is power
R is resistance

$\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}$

Hence

$\\ \bull\tt\longrightarrow R\propto V$

$\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}$

Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 0

3 years ago

We can see Objects because of

Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

7 0

4 years ago