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3 years ago

Which of the following would decrease the magnetic field around a wire?

1 answer:

4 0

The correct answer is
A. decreasing the current

In fact, the magnetic field produced by a current carrying wire is given by
 $B(r) = \frac{\mu_0 I}{2 \pi r}$
where
 $\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance from the wire at which the field is calculated

We see from the formula that the intensity of the field, B, is directly proportional to the current I, so if the current decreases, the magnetic field strength B decreases as well.

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On your first trip to Planet X you happen to take along a 180 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

lys-0071 [83]

Answer:

g_x = 3.0 m / s^2

Explanation:

Given:

- Change in length of spring [email protected] = 22.6 cm

- Time taken for 11 oscillations t = 19.0 s

Find:

- The value of gravitational free fall g_x at plant X:

Solution:

- We will assume a simple harmonic motion of the mass for which Time is:

T = 2*pi*sqrt(k / m ) ...... 1

- Sum of forces in vertical direction @equilibrium is zero:

F_net = k*x - m*g_x = 0

(k / m) = (g_x / x) .... 2

- substitute Eq 2 into Eq 1:

2*pi / T = sqrt ( g_x / x )

g_x = (2*pi / T )^2 * x

- Evaluate g_x:

g_x = (2*pi / (19 / 11) )^2 * 0.226

g_x = 3.0 m / s^2

3 0

3 years ago

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

RSB [31]

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

3 0

3 years ago

For the following statements, choose the word or words inside the parentheses that serve to make a correct statement. Each state

AnnyKZ [126]

Answer:

a) Temperatura, b) Temperature, c) Constant , d) None of these , e) Gibbs enthalpy and free energy (G)

Explanation:

a) the expression for ideal gases is PV = nRT

Temperature

b) The internal energy is E = K T

Temperature

c) S = ΔQ/T

In an isolated system ΔQ is zero, entropy is constant

Constant

d) all parameters change when changing status

None of these

e) Gibbs enthalpy and free energy

7 0

2 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$