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3 years ago

A 37.9 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.5 cm from

Physics

1 answer:

Damm [24]3 years ago

4 0

Answer:

$1.67\cdot 10^{-5} T$

Explanation:

The magnitude of the magnetic field produced by a current-carrying wire is given by the equation:

$B=\frac{\mu_0 I}{2\pi r}$

where:

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vaacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the magnetic field lines is tangential to concentric circles around the wire.

In this problem, we have:

$I=37.9 A$ is the current in the wire

$r=45.5 cm = 0.455 m$ is the distance from the wire

Solving for B, we find the magnitude of the magnetic field:

$B=\frac{\mu_0 I}{2\pi r}=\frac{(4\pi \cdot 10^{-7})(37.9)}{2\pi (0.455)}=1.67\cdot 10^{-5} T$

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A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact

Leno4ka [110]

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

6 0

3 years ago

A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su

Sedaia [141]

Answer:

$\boxed {\boxed {\sf 18 \ m/s}}$

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

$F_c= \frac{mv^2}r}$

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

$F_c$ = 85 kg*m/s²
m= 0.5 kg
r= 1.9 m

Substitute the values into the formula.

$85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}$

Isolate the variable v. First, multiply both sides by 1.9 meters.

$(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m$

$(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}$

$161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2$

Divide both sides by 0.5 kilograms.

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}$

$\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2$

$323 \ m^2/s^2 = v^2$

Take the square root of both sides of the equation.

$\sqrt {323 \ m^2/s^2} =\sqrt{ v^2$

$\sqrt {323 \ m^2/s^2} =v$

$17.9722007556 \ m/s =v$

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

$18 \ m/s =v$

The maximum speed is approximately 18 meters per second.

8 0

3 years ago

You have a string with a mass of 0.0135 kg. You stretch the string with a force of 8.29 N, giving it a length of 1.83 m. Then yo

Sphinxa [80]

The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz

Given Conditions:

mass of string; m = 0.0133 kg

Force on the string; F = 8.89 N

Length of string; L = 1.97 m

1. To find the wavelength at the fourth normal node.

At the fourth harmonic, there will be 2 nodes.

Thus, the wavelength will be;

λ = L/2

λ = 1.97/2

λ = 0.985 m

2. To find the velocity of the wave from the formula;

v = √(F/(m/L)

Plugging in the relevant values gives;

v = √(8.89/(0.0133/1.97)

v = 36.2876 m/s

Now, formula for frequency here is;

f = v/λ

f = 36.2876/0.985

f = 36.84 Hz

Read more about Harmonics of standing waves at; brainly.com/question/10274257

#SPJ4

7 0

2 years ago

The potential energy between two atoms in a particular molecule has the form U(x) = 2.1 x 8 − 5.2 x4 where the units of x are le

a_sh-v [17]

Answer:

$x\approx 0.948$

Explanation:

The correct formula for the potential energy between two atoms in a particular molecule is:

$U(x) = \frac{2.1}{x^{8}}-\frac{5.2}{x^{4}}$

Where $x$ is the distance.

According to the definitions of potential energy and work, as well as the Work-Energy Theorem and the Principle of Energy Conservation. The relation between that and related force is:

$F = -\frac{dU}{dx}$