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melisa1 [442]
3 years ago
8

Why is vinegar considered a solution?

Physics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

Vinegar is a homogenous mixture of acetic acid and water. As the mixture created has only one phase it is a solution. ... There are no chemical bonds created between water and the acid and it is possible to separate the two without breaking any chemical bonds.

Explanation:

You might be interested in
Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a
I am Lyosha [343]

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

q_1\ and\ q_2 are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

F'=\dfrac{kq^2}{r'^2}

F'=\dfrac{kq^2}{(2r)^2}

F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}

F'=\dfrac{1}{4}\times 7.35

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

3 0
3 years ago
Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?
zysi [14]
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
F= k_e \frac{ q_1 q_2}{r^2}
where k_e=8.99\cdot10^9~N m^2 C^{-2} is the Coulomb's constant, q_1=q_2=29~nC=29\cdot 10^{-9}~C is the intensity of the two charges, and r=5.5~cm=0.055~m is the distance between them.

Substituting these numbers into the equation, we get
F=2.5~10^{-3}~N

The force is repulsive, because the charges have same sign and so they repel each other.
6 0
2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
2 years ago
(ik it says physics but astronomy is a field of physics sooo) A recently discovered planet in a different solar system is locate
marusya05 [52]

The distance, to the nearest tenth, is 314.9 light-years.

<em>The given data is:</em>

A recently discovered planet is located 1.85*10^5 miles from Earth.

Now we want to transform this distance to light-years.

Remember that a light-year is defined as "the distance that the light would travel in one year".

using the relation:

distance = speed*time

The speed of light is:

speed = 6.706*10^8 mi/h

And in one year has 8760 hours, then we have:

time = 8760 h

replacing these in the equation we get:

distance = speed*time

distance = (6.706*10^8 mi/h)*(8760 h)  = 5,874,456,000,000 miles

Son one light-year is equivalent to 5,874,456,000,000 miles

1 light-year = 5,874,456,000,000 miles

So to transform a distance in miles to light-years, we just need to divide that distance by 5,874,456,000,000 miles:

The distance between the new planet and Earth was:

D = 1.85*10^15 mi = ( 1.85*10^15)/(5,874,456,000,000) = 314.9 light-years.

if you want to learn more about this, you can read:

brainly.com/question/1302132

3 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
3 years ago
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