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2 years ago

Water will expand more than ___ A. door B. juice C. air

Physics

2 answers:

Snezhnost [94]2 years ago

7 0

I think the answer is B because juice has a higher density than water

faust18 [17]2 years ago

6 0

Answer:

Its A. door because liquids expand better than solids

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Please help I'll mark brainliest!!!!

madam [21]

This question is based on the fundamental assumption of vector direction.

A vector is a physical quantity which has magnitude as well direction for its complete specification.

The magnitude of a physical quantity is simply a numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector. For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.

The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion is negative and vertically upward is positive. In case of a falling object the motion is vertically downward. So the velocity of that object is negative .

So last option is partially correct as the vector can be negative depending on our choice of co-ordinate system.

7 0

3 years ago

Read 2 more answers

HELP PLEASE...

skad [1K]

Ya because they’re are both 50 liters

7 0

3 years ago

A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to

maria [59]

Answer:

the distance between interference fringes increases

Explanation:

For double-slit interference, the distance of the m-order maximum from the centre of the distant screen is

$y=\frac{m \lambda D}{d}$

where $\lambda$ is the wavelength, D is the distance of the screen, and d the distance between the slits. The distance between two consecutive fringes (m and m+1) will be therefore

$\Delta y = \frac{(m+1) \lambda D}{d}-\frac{m \lambda D}{d}=\frac{\lambda D}{d}$

and we see that it inversely proportional to the distance between the slits, d. Therefore, when the separation between the slits decreases, the distance between the interference fringes increases.

4 0

3 years ago

A coiled spring with coils that are closely spaced then widely then closely then widely then closely, ending with a yellow line

liberstina [14]

Answer:

2, sorry I'm late but my answer is right, I just took the quiz

4 0

3 years ago

Read 2 more answers

A billiard ball moving at 5 m/s strikes another ball which is initially at rest. After the collision, the first ball moves at a

ziro4ka [17]

Answer:

The velocity of the second ball is approximately 2.588 m/s

The angle direction of the second ball is 75° counterclockwise from the horizontal

Explanation:

The initial velocity of the first billiard ball = 5 m/s

The initial velocity of the billiard ball the first billiard ball strikes = 0 m/s

The final velocity of the first billiard ball = 4.35 m/s

The final direction of motion of the first billiard ball = 30° below its original motion

For perfectly elastic collision, whereby the target is at rest initially, by conservation of momentum, we have;

m₁ × $\underset{v_1}{\rightarrow}$ = m₁· $\underset{v'_1}{\rightarrow}$ + m₂· $\underset{v'_2}{\rightarrow}$

Which gives;

m₁ × 5·i = m₁·((√3)/2×5·i - 2.5·j) + m₂· $\underset{v'_2}{\rightarrow}$

∴ m₂· $\underset{v'_2}{\rightarrow}$ = m₁ × 5·i - m₁·((√3)/2×5·i - 2.5·j)

m₂· $\underset{v'_2}{\rightarrow}$ = m₁ × 5·(1 - √3/2)·i + m₁·2.5·j = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

Therefore, given that the mass of both billiard balls are equal, we have, m₁ = m₂, which gives;

m₂· $\underset{v'_2}{\rightarrow}$ = m₁· $\underset{v'_2}{\rightarrow}$ = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

∴ $\underset{v'_2}{\rightarrow}$ = 2.5·(2 - √3)·i + 2.5·j

The magnitude of the velocity of the second ball is $\underset{v'_2}{\rightarrow}$ = √((2.5·(2 - √3))² + 2.5²) ≈ 2.588 m/s

The direction of the second ball, θ = arctan(2.5/((2.5·(2 - √3))) = 75° counterclockwise from the horizontal.

3 0

3 years ago