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3 months ago

Calculate the critical angle between glass (n = 1.90) and ice (n = 1.31)? 43° 52° 60° 75°

1 answer:

4 0

According to Snell’s Law: Where: is the index of refraction of the first medium (glass), and is the index of refraction of the second medium (ice). The angle of incidence and the angle of refraction are represented by and. The refractive index quantifies the speed of light in a medium. The critical angle is identified as the angle at which total internal reflection occurs, meaning no light passes through into another medium. This phenomenon happens only when the light is transitioning from a medium with a higher index of refraction to one with a lower index of refraction.

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A big box of sausages (30 kg) is lifted from the ground to the top shelf of the freezer. If the box is lifted at a constant spee

Yuliya22 [3333]

The work performed to elevate the box equals 515.03 J.

5 0

2 months ago

The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an

Ostrovityanka [3204]

Explanation:

The formula illustrating the relationship between resistance and temperature is as follows:

R =

$R_{o} + \alpha [T_{2} - T_{1}]$

where, R = final resistance

= initial resistance $R_{o}$

= temperature coefficient of resistivity $\alpha$

= final temperature $T_{2}$

= initial temperature $T_{1}$

Given data as follows.

$T_{1} = (20 + 273) K = 293 K$

R = 36 ohm,

= 3 ohm $T_{2}$

= 0.0045 $R_{o}$

Substituting the provided values into the above formula gives us the following.

R = $\alpha$

36 =

$R_{o} + \alpha [T_{2} - T_{1}]$

$3 + 0.0045 \times [T_{2} - 293]$

= 7626.33 K

$T_{2}$ Thus, it can be concluded that $\frac{34.3185}{0.0045}$ the temperature of the light bulb at 12.0 V is 7626.33 K.

7 0

3 months ago

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde

Maru [3345]

Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

$V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }$

Applying the natural logarithm to both sides,

$e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })$

$t = -RC ln (1-\frac{V}{V_{0} })$ (1)

Next, we can determine the capacitance using the plates' area.

C = ε₀A/d

= $\frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }$

= 18×10⁻¹²F

We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

$t = -RC ln (1-\frac{V}{V_{0} })$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

= 15 ns

5 0

2 months ago

What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?

Yuliya22 [3333]

In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

6 0

3 months ago