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cricket20 [7]
2 years ago
10

Find the mass of 250 mL of water. The density of water is 1 g/mL.​

Physics
1 answer:
emmainna [20.7K]2 years ago
4 0

Answer:

<h2>The answer is 250 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 250 mL

density = 1 g/mL

So we have

mass = 250 × 1

We have the final answer as

<h3>250 g</h3>

Hope this helps you

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What is one characteristic of a sample of matter that indicates it is a pure substance? A) It can be separated using by a physic
snow_lady [41]

D) If the composition of a sample is fixed, the sample is a pure substance.

Explanation:

A) It can be separated by using physical means, such as filtering.

Of course using physical means as filtration or crystallization we increase the purity of a compound, however sometimes impurities may remain dissolved in the solution or they may co-precipitate with the analysed sample.

B) The components in a pure substance do not have to be in definite ratios.

If the components in a substance do not have definite rations it means the the substance is not pure, it is a mixture of something.

C) If the composition of a sample varies, the sample is a pure substance.

If the composition of a sample varies, it means that different components in the substance are changing so it is a clue that the substance is not pure.

D) If the composition of a sample is fixed, the sample is a pure substance.

If the composition of a sample is fixed, and it can be determined and proven, the substance is pure.

Learn more about:

pure substances

brainly.com/question/1339017

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8 0
3 years ago
Read 2 more answers
A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0
3 years ago
What type of cooling would a scientist determine formed an igneous rock found with large crystals? Slow cooling Medium rate cool
Dmitriy789 [7]

Slower cooling engenders the growth of larger crystals in igneous rocks, thus, your answer should be slow cooling!

Hope this helped!

5 0
2 years ago
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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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