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3 years ago

9

-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the

motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

1 answer:

Nina [5.8K]3 years ago

5 0

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM given as

V(max) = ω A

ω=Angular speed

A=Amplitude

$\omega =\dfrac{3.9}{0.165}\ rad/s$

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

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A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that fina

lisov135 [29]

Answer:

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Explanation:

m = mass of air = 1 kg

T₁ = Initial temperature = 15°C

T₂ = Final temperature = 97°C

Cp = Specific heat at constant pressure = 1.005 kJ/kgk

Cv = Specific heat at constant volume = 0.718 kJ/kgk

W = Work done

Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)

ΔU = Change in internal energy

Q = W+ΔU

⇒Q = W+mCvΔT

⇒0 = W+mCvΔT

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3 years ago

Total charge is uniformly distributed on a spherical surface of radius R. The sphere is centered at the origin and spins around

Flauer [41]

Answer:

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Explanation:

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3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

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4 years ago

Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the

Leya [2.2K]

Answer: $1.298\times 10^{-5}\ N/m$

Explanation:

Given

Current in the first wire $I_1=2.2\ A$

Current in the second wire $I_2=5.9\ A$

wires are $20\ cm$ apart

Force per unit length between the current-carrying wires is

$\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

Force exerted by the wires is the same

Put the values

$\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m$

This force will be repulsive in nature as the current is flowing opposite

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3 years ago