the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of spar
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Answer:
a) 2.76x10^-9 m/s^2
b) downward
Explanation:
a) In order to do this, we need to use the newton's law of gravitation which is:
F = G.m/r²
Where:
G: Constant of gravitation (6.67x10^-11 N/m)
m: mass of the object
r: distance or radius of the object.
In this case, we have two spheres, so this law becomes:
F = G*m1*m2 / r²
Now, you are not providing a picture with this, however, this problem is very similar to one I solved before, so, I'm gonna use the same values to solve this. If, you have different values for the distance, just replace them in this procedure and should get the correct and more accurate answer.
In this case, the distance that separates sphere in P from sphere A and B is 10 cm. As both spheres have the same mass and they are at the same distance from P, we can assume that F1 = F2 so:
F1 = (6.67x10^-11) * (0.01) * (0.26) / (0.1)² = 1.73x10^-11 N
sphere at P, will roll downward to the middle of sphere A and B. The distance between these two points it's 8 cm (See picture attached)
Now, according to my data, they are exerting a force in the y axis, so:
F1y = Fy * senФ
senФ using trigonometry is:
senФ = opposite lenght / hipotenuse
senФ = 8/10 = 0.8
Replacing:
F1y = 1.73x10^-11 * 0.8 = 1.38x10^-11 N
This force it's exactly the same as F2y, so the total force is:
F = 1.38x10^-11 * 2 = 2.76x10^-11 N
Finally, using the second law of Newton:
F = m*a
Solving for a:
a = F/m
replacing:
a = 2.76x10^-11 / 0.01
a = 2.76x10^-9 m/s²
According to the picture and the value of this magnitude, it's going downward.
A) 320 count/min
B) 40 count/min
C) 80 count/min, 11400 years
Explanation:
A)
The activity of a radioactive sample is the number of decays per second in the sample.
The activity of a sample is therefore directly proportional to the number of nuclei in the sample:
where A is the activity and N the number of nuclei.
As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:
where m is the mass of the sample.
In this problem:
- When the mass is , the activity is
- When the mass is , the activity is
So we can find A2 by using the rule of three:
B)
The equation describing the activity of a radioactive sample as a function of time is:
(1)
where
is the initial activity at time t = 0
t is the time
is the decay constant, which gives the probability of decay
The decay constant can be found using the equation
where is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.
In this problem, carbon-14 has half-life of
So its decay constant is
We also know that the tree died
t = 17,100 years ago
and that the initial activity was
(value calculated in part A, corresponding to a mass of 20 g)
So, substituting into eq(1), we find the new activity:
C)
We know that a sample of living wood has an activity of
per 1 g of mass.
Here we have 5 g of mass, therefore the activity of the sample when it was living was:
Moreover, here we have a sample of 5 g, with current activity of : it means that its activity per gram of mass is
We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:
- After 1 half-life, the activity drops from 16 count/min to 8 count/min
- After 2 half-lives, the activity dropd to 4 count/min
So the age of the wood is equal to 2 half-lives, which is:
Answer:ratio of the radii of their orbits = 1.3 --- C
Explanation:
1- eV = to the kinetic energy of the electrons
and kinetic energy is given as
K.E= 1/2mv2
v = √(2E/m)----- equation 1
The force on the particles relating to the magnetic and circular motion ( centripetal force is given as
F = magnetic force = centripetal force
F= qvB = mv2/r
qvB = mv2/r
r = mv/qB ------ equation 2
We know from equation 1 that v = √(2E/m)
Therefore,
r = √(2mE)/qB------ equation 3
We can now say that the ratio of the two radii of their orbits can be calculated as
r1/r2 =(√(2mE1)/qB) /(√(2mE2)/qB
Where E1 = 500-eV and E2 = 300-eV (1- eV = to the kinetic energy of the electrons)
r1/r2 = (√(2m x500)/qB) /(√(2mx 300)/qB
Cancelling out common variables, we are left with
r1/r2 =
r1/r2= 1.29 ≈ 1.3