All categories

3 years ago

Find the gravitational force between the

Physics

1 answer:

love history [14]3 years ago

7 0

Answer:

The process is given in the pic.

I have taken the average masses so u substitute the values and solve hope it will help :)❤

You might be interested in

A car Excel accelerates from the rest to a velocity of 5 m in four seconds what is the average acceleration

MrRa [10]

Average acceleration: velocity/time

5/4= 1.25 m/s² Acceleration

Hope I helped :)

5 0

4 years ago

The rate at which work is done is also the definition of which of the following?

gregori [183]

B) Power
Because Power= Work/time

7 0

3 years ago

Which of the following statements about Boyle's law are correct? The temperature and pressure must remain constant for the law t

kvasek [131]

Answer: The temperature and the number of molecules must reamain constant for the law to apply, and as the pressure increases, the volumen decreases proportionally.

Boyle's law states that if the temperature, T, of a given mass of gas, remains constant, the Volume, V, of the gas is in inverse relation to the pressure, p; i.e.

pV = constant (for a given mass of gas, at constant T)

Then, if p increases, V decreases proportionally to keep the relation pV = constant.

8 0

3 years ago

Read 2 more answers

What is the momentum of a volleyball with a velocity of 50 mph and a mass of .5 kg?

Romashka-Z-Leto [24]

The formula for momentum is p=mv (mass multiplied by velocity), so in this problem, p=50(.5)=25=p or in other words, momentum.

6 0

3 years ago

Insert

nordsb [41]

A) 320 count/min

B) 40 count/min

C) 80 count/min, 11400 years

Explanation:

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:

$A\propto N$

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:

$A\propto m$

where m is the mass of the sample.

In this problem:

- When the mass is $m_1 = 1 g$ , the activity is $A_1=16 count/min$

- When the mass is $m_2=20 g$ , the activity is $A_2$

So we can find A2 by using the rule of three:

$\frac{A_1}{m_1}=\frac{A_2}{m_2}\\A_2=A_1 \frac{m_2}{m_1}=(16)\frac{20}{1}=320 count/min$

The equation describing the activity of a radioactive sample as a function of time is:

$A(t)= A_0 e^{-\lambda t}$ (1)

where

$A_0$ is the initial activity at time t = 0

t is the time

$\lambda$ is the decay constant, which gives the probability of decay

The decay constant can be found using the equation

$\lambda = \frac{ln2}{t_{1/2}}$

where $t_{1/2}$ 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

$t_{1/2}=5700 y$

So its decay constant is

$\lambda=\frac{ln2}{5700}=1.22\cdot 10^{-4} y^{-1}$

We also know that the tree died

t = 17,100 years ago

and that the initial activity was

$A_0 = 320 count/min$ (value calculated in part A, corresponding to a mass of 20 g)

So, substituting into eq(1), we find the new activity:

$A(17,100) = (320)e^{-(1.22\cdot 10^{-4})(17,100)}=40 count/min$

We know that a sample of living wood has an activity of

$A=16 count/min$ per 1 g of mass.

Here we have 5 g of mass, therefore the activity of the sample when it was living was:

$A_0 = A\cdot 5 = (16)(5)=80 count/min$

Moreover, here we have a sample of 5 g, with current activity of $A=20 count/min$ : it means that its activity per gram of mass is

$A'=\frac{20}{5}=4 count/min$

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:

$t=2t_{1/2}=2(5700)=11,400 y$

3 0

4 years ago