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avanturin [10]
2 years ago
14

When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.

Physics
1 answer:
Ivan2 years ago
7 0

When an object in simple harmonic motion is at its maximum displacement, its <u>acceleration</u> is also at a maximum.

<u><em>Reason</em></u><em>: The speed is zero when the simple harmonic motion is at its maximum displacement, however, the acceleration is the rate of change of velocity. The velocity reverses the direction at that point therefore its rate of change is maximum at that moment. thus the acceleration is at its maximum at this point</em>

<em />

Hope that helps!

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A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequen
MrMuchimi

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg 


The frequency of mass-spring oscillation is: 
f = (1/2π)√(k/m) 
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Then we know that k is constant for both trials, we have: 
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5 0
3 years ago
A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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6 0
2 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
uranmaximum [27]
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

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Now the total force required is:

0.0702N+0.803N=0.873N

5 0
3 years ago
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