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3 years ago

The atomic number of a nucleus decreases during which nuclear reactions?

1 answer:

5 0

A related type of beta decay actually decreases the atomic number of the nucleus when a proton becomes a neutron. Due to charge conservation, this type of beta decay involves the release of a charged particle called a “positron” that looks and acts like an electron but has a positive charge.

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How do i calculate this?

Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0

2 years ago

A car moving at a velocity of 25m/s, so how much distance it will travel in 5 seconds?

denpristay [2]

Answer:

125 meters

Explanation:

5s= s*5 so 25*5=125 m

3 0

2 years ago

Identifying Advantages of Parallel Circuits

melisa1 [442]

Answer: If one bulb goes out the other bulbs stay lit.

If there is a break in one branch of the circuit, current can still flow through the other branches.

Explanation:

3 0

2 years ago

Read 2 more answers

List at least three things that can happen as plate spread apart.

Nookie1986 [14]

Centre of Mass then axis of rotation and then moment of inertia. This was the toughest question for your level... happy to help ^_^. It was purely experimental question.

8 0

2 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

2 years ago