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4 years ago

Which characteristic best describes how nuclear decay differs from nuclear fission?

Physics

2 answers:

spin [16.1K]4 years ago

7 0

The answer is Nuclear decay is spontaneous and Nuclear fission is not.

Please mark Brainiest

kobusy [5.1K]4 years ago

6 0

Nuclear decay is spontaneous.

Nuclear fission does occur spontaneously but at a very low level in fissionable materials. The main difference between fission and nuclear decay is that fission can occur by neutron bombardment. And when the neutron source is also other fissioning atoms, a chain reaction can may happen. <span>
No chain reaction is possible with nuclear decay.
<span>That would be a much better characteristic to look at.</span></span>

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JulijaS [17]

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3 years ago

Read 2 more answers

The electric field is strongest where equipotential curves are:_______.A) closest togetherB) farthest apartC) most nearly straig

cluponka [151]

Answer:

A closest

Explanation:

This is because the electric field will be strongest or largestwhen the equipotential curves are closest together

We know that the field is

E= V/d

Where is distance and we see that d being the denominator will only make E bigger if it becomes smaller that is the curves closest

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3 years ago

Solar energy isn't usually used as the only power source for a reason because what

Schach [20]

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3 years ago

A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius

alisha [4.7K]

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

$B=\frac{\mu_oI}{2R}$ (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

$B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}$ (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

$\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A$

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

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3 years ago

Describe the conditions necessary for equilibrium on a balance.

tia_tia [17]

Answer:

Explanation:

A body under concurrent forces (concurrent forces are forces with their line of action acting at a point) is said to be in equilibrium if the sum total of the forces acting on the body is zero.

For forces acting on a balance, the sum of upward forces and the downward forced acting on the balance must be equal for the balance to be in equilibrium.

Also the sum of clockwise moment must be equal to the sum of anticlockwise moment acting on the balance.

Moment is the product of force and the perpendicular distance from the force to the pivot.

7 0

4 years ago