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Vedmedyk [2.9K]
3 years ago
12

A tall flagpole is a harmonic oscillator, flexing back and forth with a steady period. The pole rises from a base that is fixed

in the ground, but you can push it forward from chest height. To increase the amplitude of the pole's motion, you should push it forward when it isA. moving toward you.
B. as close to you as possible.
C. moving away from you.
D. as far from you as possible.
Physics
1 answer:
Alika [10]3 years ago
7 0

Answer:

C. moving away from you.

Explanation:

According to Newton's second law of motion when we multiply the mass of an object and the acceleration that the body is experiencing we get the force applied.

If the flagpole is coming towards me and I push it the velocity of the oscillation will slow as the acceleration would be in the negative direction of the motion

If the flagpole is moving away from me then the acceleration that I provide will be added to the acceleration of the oscillating flagpole which will increase the amplitude.

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How do different lense types connect to helping locate sharks through mechanical radio waves?​
Gnom [1K]

Answer:

I am not quite sure, but maybe you can find something from these websites

The second one to me is the best. Sorry if it wasn't what you needed. :(

Explanation:

https://phys.org/news/2012-11-metamaterial-lens-focuses-radio-device.html

https://animals.howstuffworks.com/fish/sharks/shark-yummy-hum1.htm

5 0
3 years ago
A man weighing 490N on earth weighs 81.7N on the moon.His mass on the moon is kg
Ymorist [56]

His mass is 50 kg . . . on the Moon, on the Earth, in the capsule rocketing between them, and on Halley's comet if he ever goes there.

6 0
3 years ago
Read 2 more answers
3 What is the displacement of a satellite when it makes a complete round along its circular path?​
kirill [66]

Answer:

0

Explanation:

The displacement is zero since it goes in a full circle and ends up where it started.

4 0
2 years ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
11. A plow pushes 100 kg of snow with 300 N of force. How much is the pile of snow<br> accelerated?
Fed [463]

Answer:

Explanation:

This is an application of Newton's second Law.

Formula

F = m * a

F = 300 N

m = 100 kg

a = ?

F = m * a

300N = 100 kg * a            Divide by 100

300N/100kg = a

a = 3 m/sec^2

3 0
2 years ago
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