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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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A telephone line hangs between two poles 14 m apart in the shape of the catenary , where and are measured in meters.
Masja [62]

Answer:

a) At x=14 the slope will be given by:

\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right).

b) Then, the angle between the line and the pole will be:

\phi=\pi - \theta

where \theta is the angle between the tangent to the catenary and the x-axis.

Explanation:

The catenary has the following general form:

y(x)==a\cosh \left({\frac {x-C_{1}}{a}}\right)+C_{2}

a) The slope at any point will be given by the derivative of y.

\frac{dy}{dx}(x)=a\sinh \left({\frac {x-C_{1}}{a}}\right)

At x=14:

\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right).

b) The angle between the tangent to the catenary and the x-axis at a given point will be given by:

\frac{dy}{dx}(x)=tan(\theta) ⇒ \theta=tan^{-1} (\frac{dy}{dx}(x))

Then, the angle between the line and the pole will be:

\phi=\pi - \theta.

5 0
4 years ago
f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
A ball is thrown horizontally. What is the ball's acceleration at its highest point
natka813 [3]
At the highest point of motion the ball comes to rest momentarily,but it is being pulled down due to the effect of gravity,so its net acceleration is downwards. So,just after that point,it starts falling downwards.
4 0
3 years ago
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves
pantera1 [17]

Answer:

It's A

Explanation:

As the waves progress through the medium, the particles they are made of move perpendicular to the direction in which the waves move. The particles do not move with the wave. So waves transmit energy but not matter as they progress through a medium.

6 0
4 years ago
The bar graph shows the population of male and female deer in a forest, measured in 10-year intervals. According to the graph, w
saul85 [17]

Answer: option D. the ratio of the population of male deer is not constant.


Explanation:


The bar graph permits to compare the results for two different populations: male and female deer in a very easy visual way.


These features are remarkable:


  • The polulation of male deer (blue bars) decrease from 1961 to 1971, then increase in the next 10 year, decrease in the next decade, and increase for the next two decades. So, its trend is erratic, with ups and downs.

        This discards the option A, which states that the population of male deer increases each decade from 1961 to 2011.


  • The population of female deer (purple or brown bars) decreases every decade.

        This discards the option B. which states that when the polulation of male deer increases, the poluplation of female deer also increases.


  • The populations never are equal, hence this discards the option C.

  • Since, one popultion increases and decreases, while the other population only decreases, you conclude that the ratio of the population of male deer to female deer is not constant, which is the option D.
4 0
4 years ago
Read 2 more answers
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