All categories

3 years ago

What is less dense water or oil steel or water helium or air or oil or water

Physics

2 answers:

Andre45 [30]3 years ago

5 0

Oil is less dense than water.

Water is less dense than steel.

Helium is less dense than air.

Leni [432]3 years ago

4 0

Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

You might be interested in

The sun is part of which type of galaxy?

jeka94

I believe the answer is D. The milky way is a spiral galaxy, You can tell just by looking at it.

4 0

3 years ago

What do "Newton's apple" and the moon have in common?

SVEN [57.7K]

Answer:

Both these motions are caused by the Gravitational force of earth.

Explanation:

Both these motions are caused by the Gravitational force of earth.

6 0

3 years ago

We know that the law of conservation of energy states that energy can not be created or destroyed. It only changes form. Conside

Pavel [41]

The conservation of energy always holds true even when not clearly observable in machines that are less than 100% efficient. More often than not a machine will suffer energy losses (e.g. consider for a cooling fan: friction between the rotating blades, drag resistance in the air the fan is pushing around, resistance in the wire, and heat radiating/conducting away from the circuitry).

7 0

4 years ago

Read 2 more answers

Se dispone de dos vasos con agua a 20 grados y queremos calentarlo hasta alcanzar 50 grados. Si el primero contiene 0,5l y el se

BigorU [14]

Answer:

The cup with 0.5L

Explanation:

To know what amount of water you take into account the specific heat of the water. The specific heat of water is:

$c_{water}=4186\frac{J}{kg\°C}$

Thus, 4186 J of energy are needed to icrease the temperature of 1 kg water in 1°C. Then, more grams of water will need more energy.

You have that one cup has 0.5 L and the other one has 750mL = 0.75L

The second cup of water will need more heat because the amount of water contained in the second cup is greater than in the first cup with 0.5L

4 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago