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dimulka [17.4K]

3 years ago

12

Suppose that atom x undergoes radioactive decay that results in atom y. If y has a larger atomic number than x, which type of ra

dioactive decay occured?

1 answer:

soldi70 [24.7K]3 years ago

4 0

Beta minus decay in which a neutron turns to a proton and B- particle

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If a 4kg Bird is pushed by the window with a force of 60 N how fast is the bird accelerate?

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Force=60N
Mass=4kg

$\\ \ast\sf\hookrightarrow Force=Mass\times Acceleration$

$\\ \ast\sf\hookrightarrow Acceleration=\dfrac{Force}{Mass}$

$\\ \ast\sf\hookrightarrow Acceleration=\dfrac{60}{4}$

$\\ \ast\sf\hookrightarrow Acceleration=15m/s^2$

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If you hold a permanent magnet the wrong way in an extremely strong magnetic f eld, its magnetization will be permanently revers

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Answer: The domain that is aligned with the applied field will grow, while the domain that is oppositely aligned to the magnetic field will shrink, this is because permanent magnets produces their own magnetic field.

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A person trapped outside during a thunderstorm should

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The movement of thermal energy from hot to cold materials is called

Fynjy0 [20]

the answer would be conduction

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Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some

erica [24]

If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.

Calculation:

Step-1:

It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.

The gauge pressure at a particular depth of ocean water is calculated as:

$P=\rho g h$

Here $\rho$ is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.

Step-2:

Now we are substituting the values to calculate the pressure at the depth of 2452-m.

$\\\begin{aligned}\\P&=\rho gh\\&=1030 (\text{ kg/m}^3)\times 9.8 (\text{ m/s}^2)\times 2452 \text{ m}\\&=24.75\times 10^6 \text{ Pa}\times\frac{1 \text{ atm}}{10.1325 \times10^4 \text{ Pa}}\\&=224.268 \text{ atm}\\\end{aligned}\\$

Learn more about gauge pressure here,

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2 years ago