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andriy [413]
3 years ago
5

The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

t A at time t is given by the formula A(t).
Physics
1 answer:
Fudgin [204]3 years ago
7 0

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer:  A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life =  5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 =  5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing   N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

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melomori [17]

Answer:

19,551 J!

Explanation:

The formula is PE = ham (h=height, a= acceleration or 9.8, m= mass)

PE = (95)(9.8)(21)

PE = 19,551 Joules

5 0
3 years ago
What is the fastest motion that can be measured in any frame of reference?
Furkat [3]

Answer: The answer is D: 300,000km/s

Explanation:

7 0
3 years ago
If an object weighs 300 N on earth, what is it’s mass on the moon?
inna [77]

Answer:

The mass of the object on the Moon (and anywhere else) is about 30.61kg. Please see more detail below.

Explanation:

Weight is the gravitational force exerted on the object and is a function of mass and gravitational acceleration:

(weight) = (mass) x (gravitational acceleration)

We are to find the mass, knowing the weight on Earth to be 300N:

(mass) = (weight on Earth) / (gravitational acceleration on Earth) = 300N / 9.8 m/s^2 = 30.61 kg

The mass of the object is 30.61kg.

The mass of the object is independent of gravity. Therefore the answer to the question "What is its mass on the Moon" is 30.61kg.

If the question were what is its weight on the Moon, the answer would be

(weight on Moon) = (mass) x (grav.accel. on Moon) = 30.61kg x 1.62 m/s^2 = 49.59N

which is about 1/6 of the object's weight on the Earth.

4 0
3 years ago
Which statement best describes the motion of an object that is being subjected to two forces of equal magnitude in opposite dire
Ludmilka [50]
The net force on the object is zero, so it has no acceleration. It moves in a straight line and at a constant speed (which may be zero).
8 0
4 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
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