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2 years ago

How many newton-meters are equal to 1600 joules

Physics

1 answer:

Salsk061 [2.6K]2 years ago

7 0

1 Joule IS 1 newton-meter.

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Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

$Thickness=\frac {\lambda}{4\times n}$

Where, n is the refractive index of the film

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {136.8\ \mu\ m}{4\times 19}$

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

7 0

2 years ago

Which of these is NOT a possible type of energy transformation?

Orlov [11]

to be franc i really think the answer is B

6 0

3 years ago

How long would it take a drag racer to increase her speed from 10.m/s to 20 m/s if her car accelerates at a uniform rate of 15 m

garik1379 [7]

It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?

5 0

3 years ago

The focal length of a lens is inversely proportional to the quantity (n-1), where n is the index of refraction of the lens of th

Ainat [17]

Answer:

46.22 cm

Explanation:

The focal refraction, fr is given by

$fr = \frac {c}{(1.572 -1)} = \frac {c}{0 .572}$

The focal red light is given by

$fv = \frac {c}{(1.605 - 1)} = \frac {c}{0.605}$

$\frac {fv}{fr} = \frac {0.572}{0 .605} = 0.945455$

$\frac {1}{fr} = \frac{1}{image} + \frac {1}{object}$ and making fr the subject we obtain

$fr = \frac {image * object}{(image + object)} = \frac {24.00 * 55} {(24.0 + 55)} = 16.70886 cm$

fv = 0.945455* 16.70886 cm = 15.79747 cm

$image = \frac {object * f} {(object - f)} = \frac {15.79747 * 24.0}{(24.0 - 15.79747)} = 46.22222 cm$

Therefore, violet image is approximately 46.22 cm

5 0

3 years ago

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu

Gemiola [76]

Answer:

$6.66\cdot 10^{-12}T$

Explanation:

The magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

$I=0.040 \mu A=4\cdot 10^{-8}A$

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

$B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T$

5 0

3 years ago