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Nookie1986 [14]
4 years ago
14

A student does an experiment with a pendulum. In the first trial, she displaces the pendulum 5 cm. In the second trial, she disp

laces the same pendulum a distance of 10 cm.
In what ways would the graphs that represent these simple harmonic motion graphs be different? In what ways would they be the same?
Physics
2 answers:
notka56 [123]4 years ago
7 0
<span>The displacement of the pendulum is directly proportional to the amplitude of the sine wave produced. The period would not change because the length of the pendulum rod remains the same.</span>
Y_Kistochka [10]4 years ago
6 0

Answer:

The graphs that represent the displacement would be different, while the graphs that represent the period would be the same

Explanation:

The displacement of the two pendulums is different. In fact, in the first trial the amplitude of the oscillation of the pendulum is 5 cm, while in the second trial the amplitude of the oscillation is 10 cm. Since the displacement is given by the equation

x(t)=A sin(\omega t)

where A is the amplitude, we see that the displacement x(t) of the two trials will be different.

On the contrary, the period of oscillation in the two trials will be the same. In fact, the period of a pendulum is given by

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum and g is the acceleration due to gravity. we see that the period depends only on the length of the pendulum, L, so since the same pendulum is used in the two trials, we can conclude that the period will be the same.

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You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what
pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

#SPJ4

8 0
2 years ago
Using the results of Question 1 that would apply if the collision were inelastic, compute (using Excel in the yellow highlighted
ch4aika [34]

Answer:

The fractional kinetic energy will be lost if the collision is inelastic. In inelastic collision, the kinetic energy is converted into other forms of energy.

The lost energy became heat and sound energy.

Explanation:

During inelastic collision, the kinetic energy of a moving object does not conserve. It changes into another form of energy such as sound energy and heat energy etc.

For example, when a moving car hit another car or wall etc, the kinetic energy is converted into sound and heat energy. This type of collision is inelastic collision.

4 0
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A car goes forward along a level road T a constant velocity the additional force needed to being the car into equilibrium is?
eimsori [14]
The additional force needed to bring the car into equilibrium is frictional force.
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2 years ago
(2.437×10⁴)(6.5411 x 10^9)/(5.37x10^6). write in scientific notation​
Musya8 [376]

Answer:

the answee is

2.968456 ×10^7

4 0
3 years ago
How to calculate the slope of position time graph?
Mashcka [7]

Answer:

The slope of a position-time graph can be calculated as:

m=\frac{\Delta y}{\Delta x}

where

\Delta y is the increment in the y-variable

\Delta x is the increment in the x-variable

We can verify that the slope of this graph is actually equal to the velocity. In fact:

\Delta y corresponds to the change in position, so it is the displacement, \Delta s

\Delta x corresponds to the change in time \Delta t, so the time interval

Therefore the slope of the graph is equal to

m=\frac{\Delta s}{\Delta t}

which corresponds to the definition of velocity.

3 0
3 years ago
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