A car with an initial speed of 23.4 km/h accelerates at a uniform rate of 0.90 m/s2 for 4.0 s.

The following equation is an example of decay? 232/90 TH---4/2 HE +228/88 RA?

DaniilM [7]

Answer:

Alpha decay

Explanation:

Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.
Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2.
Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.

5 0

2 years ago

Imagine you are holding an apple. how could you increase the potential energy of this apple?

Ray Of Light [21]

If you're holding the apple at your waist, lift it to your mouth.
Potential energy relative to any level is proportional to its height
above that level. Increase that height, and you've increased the
potential energy.

Since energy is conserved ... it never magically appears or
disappears ... you need to tell where that extra energy for the
apple came from.

It's exactly the work you did ... the force of your muscles acting
through the distance you raised the apple ... that became the
additional potential energy that the apple gained.

6 0

2 years ago

For two traveling waves, if the crest of one wave coincides with a trough of another, what type of interference occurs?

kogti [31]

When crest of one wave interferes with the trough of other wave, the amplitude of the resultant wave formed is less. Hence the type of interference is destructive interference.

3 0

3 years ago

Show all work.

lys-0071 [83]

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

$F=mg$ (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance $r$ from the centre of the Earth is given by

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

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

When the astronaut is on the Earth's surface, $r=R$ (where R is the Earth's radius), so his weight is

$F=\frac{GMm}{R^2}=640 N$

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

$r'=3R+R=4R$