A radioactive nucleus has a half-life of 5*108 years. Assuming that a sample of rock (say, in an asteroid) solidified right after the solar system formed, approximately what fraction of the radioactive element should be left in the rock today? The age of the solar system is 4.5*109 years. (No calculator is necessary, but if it helps use it.)

Answers

Answer 1
Answer:

Answer:

0.002

Explanation:

The half-life of the radioactive nucleus is related to its quantity, by the following equation:

N_(t) = N_(0)2^{-t/t_(1/2)}

Where:

N(t): is the quantity of the radioactive nucleus at time t

N₀: is the  initial quantity of the radioactive nucleus

t: is the time = 4.5x10⁹ years

t(1/2): is the half-life of the radioactive nucleus = 5x10⁸ years

(N_(t))/(N_(0)) = 2^{-4.5 \cdot 10^(9) y/5 \cdot 10^(8) y} = 2 \cdot 10^(-3)

Therefore, the fraction of the radioactive element in the rock today is 0.002.

I hope it helps you!  


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Answers

Answer:

Observed time, t = 5.58 s  

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }

t is observed time.

t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?

Answers

Answer:

The distance is 2 cm

Solution:

According to the question:

Magnetic field of Earth, B_{E} = 5.0* 10^(- 5) T

Current, I = 5.0 A

We know that the formula of magnetic field is given by:

B = \farc{\mu_(o)I}{2\pi d}

where

d = distance from current carrying wire

Now,

d = (\mu_(o)I)/(2\pi B)

d = (4\pi* 10^(- 7)* 5.0)/(2\pi* 5.0* 10^(- 5))

d = 0.02 m 2 cm

Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

If you drew magnetic field lines for this bar magnet, which statement would be true

Answers

Arrows point away from north and toward south.

and

Field lines loop around the magnet starting at the north pole and ending at the south pole.
field lines correct for future weiins

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in the +z direction exists throughout the plane of the loop. The current has the magnitude of I and it is deirected counterclockwise when observing from positive z axis.What is the magnetic force exerted by the external field on the loop?Express your answer in terms of some or all of the variables I, R, and B

Answers

Final answer:

The net magnetic force exerted by the external magnetic field on a current-carrying wire formed into a loop in a uniform magnetic field is absolutely zero since the individual forces on each section of the loop cancel each other out.

Explanation:

The force exerted by a magnetic field on a current carrying wire is given by Lorentz force law, which says that the force is equal to the cross product of the current and the magnetic field. However, in this case, where the wire is formed into a loop with current flowing in a counter-clockwise direction in presence of an external magnetic field, the individual forces on each infinitesimal section of the loop cancel each other out. Therefore, the net magnetic force exerted by the external field on the entire loop is zero.

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Final answer:

The magnetic force exerted on a current-carrying wire loop by an external magnetic field can be calculated using the equation F = I * R * B.

Explanation:

The magnetic force exerted by the external field on the current-carrying wire loop can be determined using the equation F = I * R * B. The magnetic force is equal to the product of the current, radius, and magnetic field strength. The direction of the magnetic force can be determined using the right-hand rule, where the thumb represents the direction of the current, the fingers represent the magnetic field, and the palm represents the direction of the force.

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A javelin is thrown in the air. Its height is given by ( ) 1 2 8 6 20 h x x x = − + + , where x is the horizontal distance in feet from the point at which the javelin is thrown. a. How high is the javelin when it was thrown? b. What is the maximum height of the javelin? c. How far from the thrower does the javelin strike the ground?'

Answers

The maximum height, the location on the ground and the initial vertical height of the javelin is required.

The initial height of the javelin is 6 feet.

The maximum height of the javelin is 326 feet.

The javelin strikes the ground at 160.75 feet.

The given equation is

h(x)=-(1)/(20)x^2+8x+6

where x is the horizontal distance

At x= we will get the initial vertical height.

h(0)=-(1)/(20)*0+8* 0+6\n\Rightarrow h(0)=6

Vertex of a parabola is given by

x=-(b)/(2a)=-(8)/(2* -(1)/(20))\n\Rightarrow x=80

h(80)=-(1)/(20)(80)^2+8* 80+6=326

At h(x)=0 the javelin will hit the ground

0=-(1)/(20)x^2+8x+6\n\Rightarrow -x^2+160x+120=0\n\Rightarrow x=(-160\pm √(160^2-4\left(-1\right)*120))/(2\left(-1\right))\n\Rightarrow x=-0.75,160.75

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This question is incomplete, the complete question is;

A javelin is thrown in the air. Its height is given by h(x) = -1/20x² + 8x + 6

where x is the horizontal distance in feet from the point at which the javelin is thrown.

a. How high is the javelin when it was thrown?

b. What is the maximum height of the javelin?

c. How far from the thrower does the javelin strike the ground?'

Answer:

a. height of the javelin when it was thrown is 6 ft

b. the maximum height of the javelin is 326 ft

c. distance from the thrower is 160.75 ft

Explanation:

a)

Given h(x) = -1/20x² + 8x + 6

we determine the height when x = 0

h(0) = -1/20(0)² + 8(0) + 6 = 6 ft

therefore height of the javelin when it was thrown is 6 ft

b)

to determine the maximum height of the javelin;

we find the vertex of the quadratic

so

h = - [ 8 / ( 2(-1/20) ) ] = 80

therefore

h(80) = -1/20(80)² + 8(80) + 6

= -320 + 640 + 6 = 326 ft

therefore the maximum height of the javelin is 326 ft

c)

Now the thrower is at the point ( 0,0 ) and the javelin comes down at another point ( x,0 )

this is possible by calculating h(x) = 0

⇒ -1/20x² + 8x + 6 = 0

⇒ x² - 160x - 120 = 0

⇒ x = [ -(-160) ± √( (-160)² - 4(1)(-120) ) ] /  [ 2(1) ]

x = [ 160 ± √(25600 + 480) ] / 2

so

[x = 160.75 ; x = -0.75 ]

distance cannot be Negative

therefore distance from the thrower is 160.75 ft