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3 years ago

Water and cooking oil have the same density because they're both liquids. TRUE OR FALSE

Physics

2 answers:

ki77a [65]3 years ago

5 0

Answer:

False

Explanation:

Not all liquids have the same density. We know for example that oil float on the water surface because it has less density than the water.

marissa [1.9K]3 years ago

5 0

The given statement “Water and cooking oil have the same density because they're both liquids" is false.

Answer: Option 2

<u>Explanation: </u>

It’s not necessary to have same densities if two elements are in the same state of matter. Just like solids, liquids also have different level of densities.

If you add some vegetable oil in water, you’ll see that oil will float upon the surface of the water. In fact, different types of oils and liquids float in water according to their densities that are different from one another.

Liquids have different densities because of the atomic structure they have and their respective particles mass which forms their density being the ratio of their mass and volume. Thus, liquids or air having heavier density float lower than those having lighter density.

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Why should we convert units within the metric system?

dexar [7]

We need it to calculate the how big it is, it’s mass, or even volume to get the measurements of lots of objects

6 0

3 years ago

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

3 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago

Read the scenario.

tankabanditka [31]

Answer:

distance = 6 m

Explanation:

- Distance is a scalar quantity (so, only magnitude, no direction), and it is calculated as the scalar sum of all the distances travelled by an object during its motion, regardless of the direction. So, in this problem, the distance covered by the pinecone is

d = 4 m + 2 m = 6 m

- Displacement is a vector quantity (magnitude+direction), and its magnitude is calculate as the distance in a straight line between the final position and the initial position of the object. In this case, the final position is 2 m west and the initial position is 0 m, so the displacement of the pinecone is

d = 2 m west - 0 m = 2 m west

So, a scalar quantity from this scenario is

distance = 6 m

4 0

3 years ago

Read 2 more answers