If a sample shows only one-fourth of its estimated original quantity of 14c, how old is it? think about half-lives; don't try to
1 answer:
The half-life of C-14 is approximately 5730 years. This means that the quantity contained in the sample becomes half of its original value after 5730 years. After another 5730 years, the quantity becomes 1/4 of its original value, and so on.
In our problem, the s<span>ample shows only one-fourth of its estimated original quantity, this means that its age is twice 5730 years:
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In our problem, the s<span>ample shows only one-fourth of its estimated original quantity, this means that its age is twice 5730 years:
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Answer:
to the left
Explanation:
The magnitude of the electrostatic force between two charges is given by the following equation:
where:
is the Coulomb's constant
are the magnitude of the two charges
r is the distance between the two charges
Moreover, the force is:
- Attractive if the charges have opposite sign
- Repulsive if the charges have same sign
In this problem, we have:
is the magnitude of charge 1
is the magnitude of charge 2
r = 3 m is the distance between the two charges
Substituting, we find the force on both charges:
Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Answer:
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity of the planet P1 with a period
:
(1)
Where:
is the velocity of planet P1
is the radius of the orbit of planet P1
Finding :
(2)
(3)
(4)
On the other hand, we know the gravitational force between the star S with mass
and the planet P1 with mass
is:
(5)
Where is the Gravitational Constant and its value is
In addition, the centripetal force exerted on the planet is:
(6)
Assuming this system is in equilibrium:
(7)
Substituting (5) and (6) in (7):
(8)
Finding :
(9)
(10)
Finally:
(11) This is the mass of the star S