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2 years ago

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A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside

the solenoid near its center

1 answer:

vitfil [10]2 years ago

3 0

Answer:

The value is $B = 0.0452 \ T$

Explanation:

From the question we are told that

The number of turns is N = 1000

The length is L = 50 cm = 0.50 m

The radius is r = 2.0 cm = 0.02 m

The current is I = 18.0 A

Generally the magnetic field is mathematically represented as

$B = \mu_o * \frac{N }{L} * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0$

=> $B = 0.0452 \ T$

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It requires 49 J of work to stretch an ideal very light spring from a length of 1.4 m to a length of 2.9 m. What is the value of

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Answer:

44 N/m

Explanation:

The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

The work needed to stretch a spring by <em>e</em> is given by

$W = \frac{1}{2} ke^2$

where <em>k</em> is spring constant.

$k = \dfrac{2W}{e^2}$

Using the appropriate values,

$k = \dfrac{2\times 49\text{ J}}{1.5^2\text{ m}^2} = 43.55\ldots\text{ N/m} \approx 44\text{ N/m}$

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2 years ago

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Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is

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Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

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Astronauts land on another planet and measure the density of the atmosphere on the planet surface. They measure the mass of a 50

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$1.6 kg/m^3$ is the best estimate of the density of the air on the planet.

Given:

The mass of the conical flask with stopper is 457.23 grams and the volume is $500cm^3$ .

Mass of conical flask and a stopper after removing the air is 456.43 g

To find:

The density of the air on the planet.

Solution;

Mass of the conical flask and stopper with air on the planet= 457.23 g

Mass of conical flask with a stopper and without air on the planet = 456.43 g

Mass of the air in the conical flask on the planet =m

$m = 457.23 g-456.43 g=0.8 g\\\\1 g = 0.001 kg\\\\m =0.8 g =0.8\times 0.001 kg=0.0008 kg$

The volume of the conical flask = $500 cm^3$

The volume of the air in the conical flask = $V = 500cm^3$

$1 cm^3=10^{-6} m^3\\\\V= 500cm^3= 500\times 10^{-6}m^3=0.0005 m^3$

The density of the air on the planet = d

$d=\frac{m}{V}\\\\d=\frac{0.0008 kg}{0.0005 m^3}\\\\=1.6 kg/m^3$

$1.6 kg/m^3$ is the best estimate of the density of the air on the planet.

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2 years ago

Consider the static equilibrium diagram here. What is the angle F1 must make with the horizontal?

Nataly [62]

ANSWER

$\theta=35\degree$

EXPLANATION

Since the body is in equilibrium, total upward forces must equal total downward force.

Also the net horizontal forces acting on the body must be zero.

We need to resolve $F_1$ into vertical and horizontal components.

The horizontal component is

$x=F_1\cos\theta$ .

The vertical component is

$y=F_1\sin\theta$ .

Equating the up force to the downward forces gives,

$F_1\sin\theta + 20N=60N$ .

This implies that,

$F_1\sin\theta =60N-20N$ .

$F_1\sin\theta=40N...eqn1$

Also the horizontal forces must be equal.

$F_1\cos\theta=57N...eqn2$ .

Dividing equation (1) by equation (2) gives,

$\frac{F_1\sin\theta}{F_1\cos\theta}=\frac{40}{57}$ .

$\Rightarrow \tan\theta=0.70175$

$\Rightarrow \theta=tan^{-1}(0.70175)$

$\Rightarrow \theta=35.0594$ .

Therefore the given angle that $F_1$ must make with the horizontal is approximately 35° to the nearest degree.

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An adult elephant has a mass of about 5.0 tons. An adult elephant shrew has a mass of about 50 grams. How far r from the center

Diano4ka-milaya [45]

Answer:

2023857702.507m

Explanation:

$f=\frac{GMm}{r^{2} }$

recall from newton's law of gravitation

G=gravitational constant

mshew=50g

melephant=5*10^3kg

rearth=radius of the earth 6400km or 6400000m

mearth= masss of the earth

Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2

strike out the left hand side and right hand side variables

m(shrew)/r(earth)^2 = m(elephant)/r^2

r^2 = m(elephant).r(earth)^2 / m(shrew) .........make r^2 the subject of the equation

r^2= $(5*10^{3} *(6400000)^{2} )/.05$

r^2=40960000000000

r=2023857702.507m

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2 years ago