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3 years ago

Which of these events is an example of resonance?

Physics

2 answers:

vodka [1.7K]3 years ago

8 0

An opera singer breaks a thin glass with only the use of her high frequency voice

snow_tiger [21]3 years ago

3 0

An opera singer breaks a thin glass with only the use of her high frequency voice.

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An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

Now the time required for supply of 185370 J:

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago

A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

$\theta =44.8{\circ}$

using $v^2-u^2=2as$

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration $(gsin\theta )$

s=distance traveled

$0-(1.72)^2=2(-9.81\times sin(44.8))s$

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

$0=1.72-gsin(44.8)\times t$

$t=\frac{1.72}{9.8\times sin(44.8)}$

t=0.249 s

(c)Speed of the block at bottom

$v^2-u^2=2as$

Here u=0 as it started coming downward

$v^2=2\times gsin(44.8)\times 0.214$

$v=\sqrt{2.985}$

v=1.72 m/s

3 0

3 years ago

What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency

Anvisha [2.4K]

Answer:

$1.04\times 10^{-7}$ T

Explanation:

$IP$ = Power of the bulb = 100 W

$r$ = distance from the bulb = 2.5 m

$I$ = Intensity of light at the location

Intensity of the light at the location is given as

$I = \frac{P}{4\pi r^{2}}$

$I = \frac{100}{4(3.14) (2.5)^{2}}$

$I$ = 1.28 W/m²

$B_{o}$ = maximum magnetic field

Intensity is given as

$I = \frac{B_{o}^{2}c}{2\mu _{o}}$

$1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}$

$B_{o} = 1.04\times 10^{-7}$ T

7 0

3 years ago

When we get out of the bed on a very cold morning, we feel that the air of the room is cold. But when we come back after staying

S_A_V [24]

Answer:

This is because the air outside is always cooler than the air inside, so after staying outside your body adapts to the cold air, when you come back inside, the cold air is still in you which makes the room seem warmer.

4 0

3 years ago

Determine the equivalent capacitance between points a and b.

Solnce55 [7]

Where is the figure ?

7 0

3 years ago