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Rudiy27
3 years ago
12

Question 1

Physics
1 answer:
Vesna [10]3 years ago
5 0

Answer:

1)  g = 4π² / m, 3) xaxis the  length of the pendulums and the y axis the period squared

Explanation:

a) students can approximate this system to a simple pendulum, in this case the angular velocity is

         w = √ g / l

angular velocity, frequency and period are related

         w = 2π f = 2π / T

we substitute

         T = 2π√ l / g

with this equation they can determine the value of the acceleration of gravity, for this they measure the period for various lengths of the pendulum and graph

        T² = 4π²  l / g

We graph T² vs l

where this is the equation of a line if the independent variable is y = T² and x = l

        y = (4π² / g)  l

so the slope is

         m = 4π² / g

clearing

         g = 4π² / m

where the slope can be found with the values ​​of the line not the experimental values.

2) to carry out the experiment, or the thread is attached to the sphere, the length of the pendulum that goes from the pivot point to the center of the sphere is measured with a tape measure and a small finished angle is turned or less than 10th is released, it is good to wait for the first oscillation to walk, the time of a determined number of oscillations is generally measured 10 or 20, the period is calculated

    T = t / n

a table of T² against the length is made and it is plotted with the length in the ax ax, we look for the slope and hence the acceleration of gravity

3) on the independent x-axis, the controlled variable must be plotted, which is the length of the pendulums, and on the y-axis, the dependent variable is the period squared

4) of the equation of the line

            m = 4pi2 / g

                 where it ends up reaching the floor

            g = 4pi2 / m

5) when the spring is cut, the sphere remains under the effect of gravity acceleration, the harmonic movement disappears and the sphere is in a vertical movement

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Acceleration = 192.3 m/s² (Approx.)

Explanation:

Given:

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Find:

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We know that;

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0 Kelvin

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Adjusting the amount of charge on the test charge will not change the electric field force.

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2 years ago
If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
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Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

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