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2 years ago

A certain part of a flat screen TV has a thickness of 150 nanometers. How many meters is this?

Physics

1 answer:

Bess [88]2 years ago

5 0

Answer:

1.5e-7 meters

.00000015 meters

Explanation:

.000000001 meters = 1 nanometer. Multiply that by 150 and an answer is there.

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What is the direction of the magnetic field b⃗ net at point a? Recall that the currents in the two wires have equal magnitudes.

andrew11 [14]

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that the currents in the two wires have equal magnitudes, Then:

$\bar{B_{net}} = \bar{B_1} + \bar{B_2}$

$\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}$

$\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out$

Thus; $\bar{B_{net}}$ points out of the screen at A.

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1 year ago

Samantha wants her friend to wear a bicycle helmet when they go cycling. She wants to explain how a bicycle is designed to provi

givi [52]

If you don't wear a helmet and let's say you fell off your bike, you can severely injure your head! But if you DO wear a helmet and you fell off your bike, there's about I predict a 98% chance that you won't injure but sometimes it's 100%

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7 0

2 years ago

Read 2 more answers

Which statement best explains why hydrogen’s atomic number is equal to its mass?(1 point)

igor_vitrenko [27]

Answer:

Hydrogen has one electron and one proton

3 0

1 year ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$