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3 years ago

Justify why sound waves need a medium to propagate with example??

Physics

1 answer:

Lana71 [14]3 years ago

3 0

Answer:

Sound needs a material medium for their propagation like solid, liquid or gas to travel because the molecules of solid, liquid and gases carry sound waves from one point to another. Sound cannot progress through the vacuum because the vacuum has no molecules which can vibrate and carry the sound waves.

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Car a has a mass of 5 kg and 10 m/s .cart B has a mass of 10 kg and 5 m/s .which of the following statements best compares the i

weeeeeb [17]

The two cars have same momentum

Explanation:

The momentum of an object is given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

For the car A in this problem,

m = 5 kg

v = 10 m/s

So its momentum is

$p_A = (5 kg)(10 m/s) = 50 kg m/s$

For car B we have,

m = 10 kg

v = 5 m/s

So its momentum is

$p_B = (10 kg)(5 m/s)=50 kg m/s$

Therefore, the two cars have same momentum.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

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#LearnwithBrainly

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2 years ago

what is the property of an object due to its mass by which it resists any change in its position unless overcome by force ?

UkoKoshka [18]

Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

hope this helps

5 0

2 years ago

If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he

forsale [732]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q= mC_s \Delta T$
where m is the mass of the substance, Cs is its specific heat capacity and $\Delta T$ is the increase of temperature.

If we re-arrange the formula, we get
$C_s = \frac{Q}{m \Delta T}$
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
$C_s = \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C$

6 0

3 years ago

Read 2 more answers

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago

A student uses the right-hand rule as shown.

m_a_m_a [10]

Answer: O:right

Explanation: In this case you place your finger on the current, and your fingers should curl showing the way, you must use your right hand in this case, otherwise that would mean the fingers on your left would bend way back, and snap off, (Not really lol, just saying)

5 0

2 years ago