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Dimas [21]
2 years ago
10

If we double only the amplitude of a vibrating ideal mass-and-spring system, the mechanical energy of the system:

Physics
1 answer:
Lelechka [254]2 years ago
8 0

Answer:

D. increases by a factor of 4.

Explanation:

General equation of SHM

Lets taken the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

V=\dfrac{dx}{dt}

V= A ω cosωt

The mechanical energy of spring mass system

U=\dfrac{1}{2}KA^2

K=Spring constant

Now when Amplitude A become 2 times then the mechanical energy will become 4 times.

Therefore the answer is D.

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Two sinusoidal waves travel along the same string. They have the same wavelength and frequency. Their amplitudes are ym1 = 2.5 m
Nimfa-mama [501]

Answer:

0.5 m

Explanation:

Givens:

ym1 = 2.5 mm

ym2 = 4.5 mm

Ф_1=π / 4

Ф_2=π / 2

We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is

Ym = (ym1 + ym2)cos(Ф_2/2)

By substitution we have  

Ym= (0.025 + 0.045)cos(π/4) = 0.496 m

The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore  

Ym^2=(ym1^2+ym2^2)

So we have Ym=√0.025^2+0.045^2

                         = 0.5 m

7 0
2 years ago
If two waves with identical crests and troughs meet, what is happening?
Nastasia [14]
<span>If two waves with identical crests and troughs meet, what is happening?
</span>C. Constructive interference is occurring. 
3 0
2 years ago
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A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an
Cloud [144]

Answer:

1.36

Explanation:

n_{air} = Index of refraction of air = 1

n_{plastic} = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

n_{air} Sini = n_{plastic} Sinr

(1) SIn32.0 = n_{plastic} Sin23.0

n_{plastic} = 1.36

5 0
2 years ago
A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the
kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0
1 year ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602  °C

5 0
2 years ago
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