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Mademuasel [1]
3 years ago
5

Is it true of false that the slope of a line equals the average speed of the object.

Physics
1 answer:
kupik [55]3 years ago
8 0
IF the object's motion is graphed, and IF time is plotted
along the x-axis and the object's position is plotted along
the y-axis, then the slope of the graph at every point is the
object's speed at that instant of time.
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_____ in the uterus has to occur for a woman to confirm her suspicions that she is pregnant.
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Conception is the correct answer
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Fast and safe heart rate for workouts is called muscular strength? True or false
Vsevolod [243]

Answer:

False

Explanation:

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The mass of a proton is 1833 times larger than the mass of an electron. When a proton an electron interact with each other the o
Ad libitum [116K]

B. Exactly the same as the electric force of the electron on the proton.

<u>Explanation:</u>

Even if the mass of proton is increased or decreased, the force between electron and proton will remain the same because force is dependent on the charge of the object and distance between them. The force between the charges is independent of their masses. So, even if the mass of a proton is 1833 times larger than the mass of an electron, the force between them will be same.

According to Coulomb's law:

F = k\frac{q_1q_2}{r^2}

where,

F is the force

q₁ and q₂ are the charges

r is the distance between the charges

8 0
3 years ago
Insert
nordsb [41]

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:

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

B)

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

C)

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
The burning of wood in a campfire is which type of reaction?
grin007 [14]
The answer is C
Hope this help
4 0
3 years ago
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