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Juli2301 [7.4K]
3 years ago
15

The ability to move or change an object or what a wave carries

Physics
1 answer:
Sever21 [200]3 years ago
5 0

Answer: The ability to move or change an object or what a wave carries is called Energy

Explanation: Waves are disturbances in physical quantities. Example of waves are light waves, sound waves, or transverse oscillations of a string. These disturbances use energy to create and propagate, for it to move the constituent particles or change the electric or magnetic fields. Therefore, power of a wave is therefore, energy transported divided by unit time caused by the oscillations of a particular wave. The derivation of a formula for the power depends on the medium -- for light waves, the power is measured by the pointing vector, whereas for oscillations on a string, the power can be computed directly by balancing forces through the application of newton law. However, for all types of waves, the formula and physical meaning of the power takes similar forms, typically depending on the square amplitude of the waves among other factors.

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find the resistance of a resistor connected to a 3v voltmeter and a 3A ammeter, resistance box along with cells of EMF 3V​
yarga [219]

Answer:

0

Explanation:

Emf E= V+ terminal V

3 = 3R + 3

R=0

5 0
3 years ago
People, especially adults, are inactive because of "passive" modes of transportation.
enot [183]

The drop in physical activity is partly due to inaction during leisure time and sedentary behaviour on the job and at home. Likewise, an increase in the use of "passive" modes of transportation also contributes to insufficient physical activity.

3 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
Complete the passage to describe the relationship between kinetic energy, internal energy, thermal energy, and
butalik [34]

Answer:

Increases

Increases

Increases

Explanation:

I don't know if you answered your own question but I'll just answer this for others confused ahh

4 0
2 years ago
Read 2 more answers
Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0
2 years ago
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