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wel
3 years ago
7

When an unstable radioactive atom degrades, what happens to it?

Physics
1 answer:
malfutka [58]3 years ago
7 0

it depends on the atom . but in some cases it may explode

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A bowling ball of mass m=1.7kg is launched from a spring compressed by a distance d=0.31m at an angle of theta=37 measured from
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Answer:

k = 1 700.7 N/m

v0 = 9.8 m/s^2

Explanation:

Hello!

We can answer this question using conservation of energy.

The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.

When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.

PS = (1/2) k x^2  <em>where x is the compresion or elongation of the spring</em>

PG = mgh

a)

Since energy must be conserved and we are neglecting any energy loss:

PS = PG

Solving for k

k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)

k = 1 700.7 N/m

b)

Since the potential energy of the spring transfors to kinetic energy of the ball we have that:

PS = KE

that is:

(1/2) k x^2 = (1/2) m v0^2

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v0 = 9.8 m/s^2

8 0
3 years ago
Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 tim
Oksi-84 [34.3K]

Answer:

1.117935:1

Explanation:

Since the wires are of the same material, they will have the same resistivity \rho.

The cross-sectional area of the of a wire is given by;

A=\pi\frac{d^2}{4}................(1)

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

\rho=\frac{RA}{l}..................(2)

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)

Given;

l_1=1.35l_2

\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)

We then use equation (1) to fine the ratio of the area A_1 to A_2

bearing in mind that d_1=0.91d_2

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

\frac{R_1}{R_2}= 1.35*0.8281=1.117935

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