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Alexxx [7]
3 years ago
14

The displacement in simple harmonic motion is a maximum when the 1. velocity is a maximum. 2. kinetic energy is a maximum. 3. ve

locity is zero. 4. acceleration is zero. 5. linear momentum is a maximum
Physics
1 answer:
dlinn [17]3 years ago
5 0

Answer:

3. velocity is zero.

Explanation:

The velocity of a simple harmonic motion is given by

v = \omega\sqrt{A^2-x^2}

Here, <em>ω</em> is the angular velocity, <em>A</em> is the amplitude (or maximum displacement from the equilibrium point) and <em>x</em> is the displacement at any time.

At maximum displacement, <em>x </em>=<em> A</em>.<em> </em>Then

v = \omega\sqrt{A^2-A^2} = 0

Therefore, at maximum displacement, velocity is 0.

Practically, this can be observed in a simple pendulum. As it approaches the maximum displacement, its velocity reduces. It becomes zero at this point and then reverses as the pendulum changes course. Then the velocity begins to increase. It becomes maximum at the equilibrium point but once past that, the velocity begins to reduce as it approaches the other amplitude.

For acceleration,

a = -\omega^2x

It follows that at maximum displacement, the acceleration is a maximum. The negative sign indicates that it is in an opposite direction to the displacement. Both kinetic energy (\frac{1}{2}mv^2) and linear momentum (mv) are proportional to velocity; they are therefore both zero at the maximum displacement.

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A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1
Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = \rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = \frac{\rho dx}{A}

dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}

Taking the integral of both sides;we have:

\int\limits^R_0  dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx

R = 3.185 [x + \frac {x^3}{3}}]^2__0

R = 3.185 [2 + \frac {2^3}{3}}]

R = 14.863 Ω

Since V = IR

I = \frac{V}{R}

I = \frac{17}{14.863}

I = 1.144 A

∴  the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0
3 years ago
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
ki77a [65]

Answer:

The lowest possible frequency of sound for which this is possible is 1307.69 Hz

Explanation:

From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.

First, we will determine his distance from the second speaker using the Pythagorean theorem

l₂ = √(2.00²+5.00²)

l₂ = √4+25

l₂ = √29

l₂ = 5.39 m

Hence, the path difference is

ΔL = l₂ - l₁

ΔL = 5.39 m - 5.00 m

ΔL = 0.39 m

From the formula for destructive interference

ΔL = (n+1/2)λ

where n is any integer and λ is the wavelength

n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.

Then,

0.39 = (1+ 1/2)λ

0.39 = (3/2)λ

0.39 = 1.5λ

∴ λ = 0.39/1.5

λ = 0.26 m

From

v = fλ

f = v/λ

f = 340 / 0.26

f = 1307.69 Hz

Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.

5 0
3 years ago
A student performs an activity to study how electric current flows in a circuit. The student constructs two different circuits,
natta225 [31]

Native_Americans_in_the_United_State0000000000000000000000000

Explanation:

8 0
3 years ago
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Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll
Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

Q = \dfrac{4mME }{(m+M)^2}

However; from the total stopping power & power loss of the electron;

\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  =\dfrac{82 \times 1.9}{800}

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

4 0
3 years ago
What is the mass of the object if it has a density of 657 g/mL and a volume of 32 mL?<br> Show work!
Luden [163]

Answer:

The answer is 21024g/mL

Explanation:

Multiply 657 by 32:

 657

×  32

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

21024

   ↳            21024 g/mL

3 0
3 years ago
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