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3 years ago

A disturbance sends ripples across water in a tub. These ripples are an example of a rarefaction

1 answer:

3 0

The appropriate response is a Surface wave. It is a seismic wave that goes over the surface of the Earth rather than through it. Surface waves as a rule have bigger amplitudes and longer wavelengths than body waves, and they travel more gradually than body waves do. Adore waves and Rayleigh waves are sorts of surface waves.

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What unit can be used to measure length or distance

strojnjashka [21]

Answer:

The most common units that we use to measure length in the metric system are the millimeter, centimeter, meter, and kilometer.

Explanation:

7 0

3 years ago

In a closed system, the loss of momentum of one object ________ the gain in momentum of another object.

Virty [35]

In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁ = - (P'₂ - P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

8 0

3 years ago

Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west

Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

$v_B=+5\ m/s$ is the velocity of Bill with respect to Amy (which is stationary)

$v_c=15\ m/s$ is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

$v_B=+5\ m/s$

Therefore, Carlos velocity in Bill's reference frame will be

$v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s$

So, the magnitude is 20 m/s and the direction is westward (negative sign).

7 0

3 years ago

A mass moves back and forth in simple harmonic motion with amplitude A and period T.

Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

$1 T : 4 A = t : 2 A$

and solving for t we find

$t=\frac{(1T)(2 A)}{4A}=0.5 T$

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

$1 T : 4 A = t : 5 A$

And by solving for t, we find

$t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T$

6 0

3 years ago

when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2

Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂ = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0

3 years ago