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insens350 [35]
3 years ago
5

How is radioactivity produced

Physics
1 answer:
aev [14]3 years ago
8 0

Explanation:

It is a physical observed phenomenon, where an atom is emitting type of particles such as, alpha or beta followed by gamma radiations, which change the radioactive atom.

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The density of aluminum is 2.70 g/ml. A piece of aluminum foil has a mass of 44 g. What is the volume of this piece of aluminum
anyanavicka [17]

Answer:

C) 16.3 ml

Explanation:

Density is equal to the ratio between the mass of an object and its volume:

d=\frac{m}{V}

where

m is the mass

V is the volume

In our problem, we know:

- density of aluminium: d=2.70 g/mL

- mass of the aluminium foil: m=44 g

So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

V=\frac{m}{d}=\frac{44 g}{2.70 g/mL}=16.3 mL

7 0
3 years ago
List two of the number of strategies you have learned
jolli1 [7]

Answer:

what?

Explanation:

does that means also i learned Ratios and multiplication i keared them like 5 years ago?

7 0
2 years ago
A hydraulic machine can be used to lift extremely heavy objects. Why is the fluid in the hydraulic machine a liquid rather than
Sauron [17]
Hi,

The force that acts on hydraulic machine is heavy therefore the content must be something that cannot be compressed by that kind of force, the gas can easily be compressed while a liquid is nearly impossible to.
4 0
3 years ago
Read 2 more answers
A race car starts from rest and accelerates uniformly to 69 mph in 4.5 s.
Marizza181 [45]
For this question we should apply
a = v^2 - u^2 by t
a = 69 - 0 by 4.5
a = 69 by 4.5
a = 15.33
a = 6.85 m/s^2

If the answer in option is near to answer then , you can mark it as correct.
.:. The acceleration is 6.9 m/s^2
5 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
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