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3 years ago

6. Thorium-234 decays by giving off one particle, and the decay can be represented by the following nuclear equation:

Physics

1 answer:

Nady [450]3 years ago

8 0

23892U→l23490Th+42He

Explanation:

Uranium-238 produces thorium-234 by alpha decay.

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, uranium-238 decays through α-particle emission to form thorium-234 according to the equation:

23892U→l23490Th+42He

Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation.

Also, the sum of the superscripts (masses) is the same on each side of the equation.

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. A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and

Lady bird [3.3K]

Answer:

The resulting magnetic force on the wire is -1.2kN

Explanation:

The magnetic force on a current carrying wire of length 'L' with current 'I' in a magnetic field B is

F = I (L*B)

Finding (L * B) , where L = (2, 0, 0)m , B = (30, -40, 0)

L x B = $\left[\begin{array}{ccc}i&j&k\\2&0&0\\30&-40&0\end{array}\right]$ = (0, 0, -80)

we can now solve

F = I (L x B) = I (-80)

F = -1200 kmN

F = -1200 kN * 10⁻³

F = -1.2kN

6 0

3 years ago

A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.

yulyashka [42]

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization, $M=5.3\times 10^3\ A/m$

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

$M=\dfrac{\mu}{V}$

$\mu=M\times \pi r^2\times l$

Where

r is the radius of rod, r = 0.0075 m

$\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m$

$\mu=0.0683\ J/T$

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i

laiz [17]

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==> x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==> v ≈ 1.43 m/s

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0

2 years ago

The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are

Anna007 [38]

Answer:Negatively charged particle called Free Electrons

Explanation:

Current is the flow of charged particles called Free electrons. Electrons are free to move from one atom to another and we call them a sea of de-localized electrons. In absence of any externally applied emf, these electrons are randomly moving but with the onset of emf, these electrons flow in a particular direction.

6 0

3 years ago

1) A pendulum is configured to have a period of 2 seconds.

vazorg [7]

"The" (and any subsequent words) was ignored because we limit queries to 32 words.

3 0

3 years ago