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3 years ago

How to convert 1 meter into decimeters, milimeters, kilometers, micrometers etc..

Physics

1 answer:

zzz [600]3 years ago

8 0

Answer:

●1m to KM

Divided by 1000

●1m to Milli meter

Multiple by 1000

●1 decimeter =0.1 m

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A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastwar

lara31 [8.8K]

Answer:

The values is $B = 3.2 *10^{-8} \ T$

The direction is out of the plane

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 9.6 \ V/m$

The magnitude of the magnetic field is mathematically represented as

$B = \frac{E}{c}$

where c is the speed of light with value

$B = \frac{ 9.6}{3.0 *10^{8}}$

$B = 3.2 *10^{-8} \ T$

Given that the direction off the electromagnetic wave( c ) is northward(y-plane ) and the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page (z-plane )

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3 years ago

17. Which of the following is NOT a testable

Artemon [7]

Answer:

c. testing student opinions

Explanation:

opinions aren't factual and they would not aide an experiment if it wasn't for a social experiment.

7 0

3 years ago

PLEASE HELP URGENT: Please see attachment for problem! AP physics projectile question. I'd REALLY appreciate any help.

kiruha [24]

I’m so sorry, I need more information. Good luck and I’m sorry I couldn’t help you :(

4 0

3 years ago

A solid uniform cylinder of mass 4.1 kg and radius 0.057 m rolls without slipping at a speed of 0.79 m/s. What is the cylinder’s

Amiraneli [1.4K]

Answer:

The cylinder’s total kinetic energy is 1.918 J.

Explanation:

Given that,

Mass = 4.1 kg

Radius = 0.057 m

Speed = 0.79 m/s

We need to calculate the linear kinetic energy

Using formula of linear kinetic energy

$K.E_{l}=\dfrac{1}{2}mv^2$

$K.E_{l}=\dfrac{1}{2}\times4.1\times(0.79)^2$

$K.E_{l}=1.279\ J$

We need to calculate the rotational kinetic energy

$K.E_{r}=\dfrac{1}{2}\times I\omega^2$

$K.E_{r}=\dfrac{1}{2}\times\dfrac{1}{2}\times mr^2\times(\dfrac{v}{r})^2$

$K.E_{r}=\dfrac{1}{4}\times m\times v^2$

$K.E_{r}=\dfrac{1}{4}\times4.1\times(0.79)^2$

$K.E_{r}=0.639\ J$

The total kinetic energy is given by

$K.E=K.E_{l}+K.E_{r}$

$K.E=1.279+0.639$

$K.E=1.918\ J$

Hence, The cylinder’s total kinetic energy is 1.918 J.

8 0

3 years ago

Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu

ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

= 223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0

3 years ago