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Viefleur [7K]
2 years ago
6

"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five t

imes, being careful to release the ball from the same starting Point A and find the average horizontal distance traveled to be 2.0 m. One student suggests they use a stopwatch to find the time the ball is in the air whereas another student suggests they use a meter stick. Given these materials, describe the procedure students should follow to minimize error and calculate the speed of the ball as it leaves the ramp.
Physics
1 answer:
romanna [79]2 years ago
6 0

Answer:

Speed of ball as it leaves ramp is 2 m/s

Explanation:

From the question it is given that the average distance traveled by ball from point A to B is 2.0 m and the time taken is 1sec

one student is using stopwatch to calculate the time taken by ball and another student is using meter stick to calculate the distance traveled

since speed of ball is given by v = \frac{distance}{time}  thus,

the speed = 2 m/s

now for minimizing the error of speed it is necessary to record the readings by single students at-least 5 times and take average

by doing this, the error of speed calculation will be minimum as the it decreases the error due to random error of system caused by taking the reading  by different students

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2 years ago
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta
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Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

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ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

F_e=k\frac{qq'}{r^2}             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

a_c=\frac{v^2}{r}                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}

Finally, you replace the values of all parameters in the previous expression:

r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m

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5 0
3 years ago
A laser beam is incident on two slits with a separation of 0.230 mm, and a screen is placed 4.75 m from the slits. If the bright
Mariulka [41]

Answer:

7.55\times 10^{-7} m

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We know that

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\lambda=\frac{1.56\times 10^{-2}\times 0.23\times 10^{-3}}{4.75}

\lambda=7.55\times 10^{-7} m

4 0
2 years ago
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