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The weight of a box is found to be 30 N. What is the approximate mass of the box?

KengaRu [80]

The mass of the box would be 30!

3 0

3 years ago

According to Kepler's Second Law the radius vector drawn from the Sun to a planet Multiple Choice is the same for all planets. s

Mademuasel [1]

Answer:

sweeps out equal areas in equal times.

Explanation:

As we know that there is no torque due to Sun on the planets revolving about the sun

so we will have

$\tau_{net} = 0$

now we have

$\frac{dL}{dt}= 0$

now we also know that

$Area = \frac{1}{2}r^2d\theta$

so rate of change in area is given as

$\frac{dA}{dt} = \frac{1}{2}r^2\frac{d\theta}{dt}$

so we will have

$\frac{dA}{dt} = \frac{1}{2}r^2\omega$

$\frac{dA}{dt} = \frac{L}{2m}$

since angular momentum and mass is constant here so

all planets sweeps out equal areas in equal times.

4 0

3 years ago

Which of the following could possibly be predicted? Earthquakes Landslides Foreshocks Volcanic eruptions

NeTakaya

Answer:

Volcanic Eruptions

Explanation:

The volcano can start showing signs that it may be about to explode.

6 0

3 years ago

The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne

aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ $V_{H} =\frac{I}{nqt} B$

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ $\frac{V}{T}$

b = —0.02 μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH = $100X10^{-6}$ V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

$a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm$

To learn more about the hall voltage , Visit: brainly.com/question/19130911

#SPJ4

8 0

2 years ago

An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta

icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

$v=\sqrt{\frac{T}{\mu} }$

where T= 200 N is tension in the string , $\mu$ =1.0 g/m is the linear mass density

$v=\sqrt{\frac{200}{1\times 10^{-3} }$

$v=447.2 m/s$

Wavelength of the wave in the string is

$\lambda =2L=2\times 0.8=1.6 m$

The frequency is

$f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz$

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

$\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m$

8 0

3 years ago