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amm1812
2 years ago
8

What happens in a nucleus during beta decay

Physics
1 answer:
Jet001 [13]2 years ago
7 0

The number of protons plus the number of neutrons decreases

Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other. In beta minus decay, a neutron decays into a proton, an electron, and an antineutrino: n Æ p + e - +.

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Which statement is true about the force of gravity?
kumpel [21]

Answer:

The moon's gravity pulls the Earth to make tides.

Explanation:

The Moons Gravity Pulls On The Earth With Different  Strenght Making High Tide And Low Tide.

Hope This Helps!

4 0
2 years ago
Which of these events indicate that a chemical reaction might have occurred?
blagie [28]
Reactions occur when two or more molecules interact and the molecules change. Bonds between atoms are broken and created to form new molecules. That's it.
7 0
3 years ago
A balloon is ascending at 12.4m/s at a height of 81.3m above the ground when a package is dropped. a) How long did it take to re
tankabanditka [31]

Answer:

3secs

Explanation:

Given the following parameters

height H= 81.3m

Velocity v = 12.4m/s

Required

Time it take to reach the ground

Using the equation of motion

H = ut+1/2gt²

81.3 = 12.4t + 1/2(9.8)t²

81.3 = 12.4t + 4.9t²

4.9t² + 12.4t - 81.3 = 0

Using the general formula to find t

t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)

t = -12.4±√153.76+1593.48/2(4.9)

t = -12.4±√1747.24/9.8

t = -12.4+41.8/9.8

t = 29.4/9.8

t = 3secs

Hence it took 3secs to reach the ground

5 0
3 years ago
In nature, where do fusion reactions take place?
hjlf
Fusion reactions happen in stars 
3 0
2 years ago
Read 2 more answers
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What
AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, \alpha=3.3 rad/s^2

Diameter of the wheel, d=21 cm

Radius of wheel, r=\frac{d}{2}=\frac{21}{2} cm

Radius of wheel, r=\frac{21\times 10^{-2}}{2} m

1m=100 cm

Magnitude of total linear acceleration, a=1.7 m/s^2

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,a_t=\alpha r

a_t=3.3\times \frac{21\times 10^{-2}}{2}

a_t=34.65\times 10^{-2}m/s^2

Radial acceleration,a_r=\frac{v^2}{r}

We know that

a=\sqrt{a^2_t+a^2_r}

Using the formula

1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}

Squaring on both sides

we get

2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}

\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}

v^4=r^2\times 2.7699

v^4=(10.5\times 10^{-2})^2\times 2.7699

v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}

v=0.418 m/s

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0
2 years ago
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