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kari74 [83]
1 year ago
15

10. Complete each of the following radioactive decay equations.

Physics
1 answer:
alukav5142 [94]1 year ago
6 0

1. Alpha

Number next to Th is 231

(because 231 + 4 = 235)

The other number below the 4 is 2 - helium

(because 90 + 2 = 92)

2. Beta

Missing numbers are zero and one. Same as the one below it.

3. Beta

11 is the number above the C

(because 11 + 0 = 11)

5 is the number next to the B

(because 5 + 1 = 6)

Basically, the missing numbers must balance on both side of the equation; creating the elements you've started of with, meaning if you where to add the right side of the equation up - you should end up with the same protons and mass numbers you began with.

Hope this helps!

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A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
A bowling ball with a mass of 8 kg is moving at a speed of 5m/s. What is its kinetic energy
dolphi86 [110]

Answer:The formula for kinetic energy is

(1/2) M V^2.

With M in kg and V in m/s, the answer will be in Joules.

K.E = 40joule

Explanation:

4 0
3 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kg when weighed in air. The density of
sasho [114]

Answer:

m = 4.5021 kg

Explanation:

given,

Apparent mass of aluminium = 4.5 kg

density of air = 1.29 kg/m³

density of aluminium = 2.7 x 10⁷ kg/m³

true mass of the aluminium = ?

Weight in Vacuum

W = m g

W = ρV g

Air buoyancy acting on aluminium

B = ρ₀V g

Volume is the same in both cases since the volume of the aluminum

displaces an equal amount of volume air.

Apparent weight:

ρV g − ρ₀V g = 4.5 g

ρV − ρ₀V = 4.5

V = \dfrac{4.5}{\rho - \rho_0}

m = ρV

m = \dfrac{4.5\times \rho}{\rho - \rho_0}

m = \dfrac{4.5\times 2700}{2700 - 1.29}

m = 4.5021 kg

4 0
3 years ago
What is the relationship in which the ratio of the manipulated variable and the responding variable is constant?
son4ous [18]
That's a 'direct' or 'proportional' variation.

The variables are related by the equation      Y = K X .

That constant ratio is the number    ' K ' .

5 0
3 years ago
The half-lives for the radioactive decay of various elements are listed below.
myrzilka [38]

Answer:

D ko alam pasensya ka na ha

4 0
3 years ago
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