Answer:
0.002
Explanation:
The half-life of the radioactive nucleus is related to its quantity, by the following equation:
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
Therefore, the fraction of the radioactive element in the rock today is 0.002.
I hope it helps you!
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 is observed time.
So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?
Answer:
The distance is 2 cm
Solution:
According to the question:
Magnetic field of Earth, B_{E} =
Current, I = 5.0 A
We know that the formula of magnetic field is given by:
where
d = distance from current carrying wire
Now,
d = 0.02 m 2 cm
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves
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.
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.
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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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.
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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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
where is the horizontal distance
At we will get the initial vertical height.
Vertex of a parabola is given by
At the javelin will hit the ground
Learn more about parabolas from:
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