In the experiment, measuring the total time for 20 complete revolutions and dividing it by 20 to obtain the period of rotation is done to reduce errors and improve the accuracy of the measurement.
Measuring the time for one complete revolution directly can be subject to human reaction time and potential errors in starting and stopping the stopwatch precisely at the beginning and end of each revolution. These errors can accumulate and affect the accuracy of the measurement.
By measuring the total time for 20 complete revolutions and then dividing it by 20, we are essentially averaging out these potential errors over multiple revolutions. This helps to minimize the impact of any individual timing error and provides a more reliable and accurate measurement of the period of rotation.
Additionally, by taking multiple measurements (in this case, 20), we increase the sample size and reduce the influence of outliers or irregularities in any individual measurement. This improves the overall precision and reliability of the calculated period.
Therefore, measuring the total time for multiple revolutions and dividing by the number of revolutions allows for a more accurate determination of the period of rotation in the experiment.
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A particle with a charge of 5nC has a distance of 0. 5m away from a charge of 9. 5nC. What is its electric potential energy?
The electric potential energy of the particle with a charge of 5nC, located 0.5m away from a charge of 9.5nC, is 1.9 J.
To calculate the electric potential energy, we can use the formula:
Electric potential energy = (k * q1 * q2) / r
Where:
k is the electrostatic constant (9 x 10^9 N m^2/C^2),
q1 and q2 are the charges of the two particles (in this case, 5nC and 9.5nC, respectively),
r is the distance between the charges (0.5m).
Substituting the given values into the formula:
Electric potential energy = (9 x 10^9 N m^2/C^2) * (5 x 10^-9 C) * (9.5 x 10^-9 C) / 0.5m
Calculating the expression:
Electric potential energy ≈ 1.9 J
Therefore, the electric potential energy of the particle is approximately 1.9 Joules.
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A projectile is launched horizontally from a height of 8. 0 m. The projectile travels 6. 5 m before hitting the ground. The velocity of the projectile the moment it was launched, rounded to the nearest hundredth, is m/s.
The initial velocity of a projectile launched horizontally can be calculated using the equation of distance covered horizontally (x) = Initial velocity (u) Time of flight (t). The horizontal component of the initial velocity can be determined by x = u t, t = 1.63 s, x = 6.5 mu = x / t = 6.5 m / 1.63 su = 3.99 m/s 4.00 m/s.
The initial velocity of the projectile that was launched horizontally can be calculated using the equation below: Distance covered horizontally (x) = Initial velocity (u) × Time of flight (t) where, Time of flight (t) can be found using the formula below: t = [2 × vertical height (h)] / g where ,g is the acceleration due to gravity = 9.8 m/s².The vertical height (h) of the projectile is 8.0 m. So the time of flight of the projectile will bet = [2 × 8.0 m] / 9.8 m/s²t = 1.63 s Therefore, the horizontal component of the projectile’s initial velocity can be determined by: x = u × tt = 1.63 s, x = 6.5 mu = x / t = 6.5 m / 1.63 su = 3.99 m/s ≈ 4.00 m/s. So, the projectile was launched horizontally with a velocity of 4.00 m/s (rounded to the nearest hundredth).Content loaded: The term “content loaded” is used to indicate that the contents of a webpage or app have finished loading and are ready for viewing or use.
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A 5-kg object is moving to the right at 4 m/s and collides with another object moving to the left at 5 m/s. The objects collide and stick together. After the collision, the combined object:
After the collision, the two objects stick together and move as one. Their total mass is m1 + m2 = 5 kg + m2.
How to determine the effect of the collisionIn this case, we can apply the principle of conservation of linear momentum
The initial momentum of the first object (P1_initial) is given by its mass (m1) times its velocity (v1), which is [tex]5 kg * 4 m/s = 20 kg*m/s.[/tex]
Therefore, the total initial momentum [tex](P_{total_initial}) is P1_{initial} + P2_{initial} = 20 kg*m/s - m2 * 5 m/s.[/tex]
After the collision, the two objects stick together and move as one.
Their total mass is m1 + m2 = 5 kg + m2.
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A stevedore slides a crate along a dock with a 50 kg horizontal force of 175 N. The opposing force of friction is 120 N. If started from rest, what is the crates’s final velocity after 0.5s?
To determine the crate's final velocity after 0.5 seconds, we can use the concept of Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
In this scenario, the stevedore applies a horizontal force of 175 N to move the crate along the dock. However, there is also an opposing force of friction acting in the opposite direction, which has a magnitude of 120 N. The net force is the difference between these two forces, so we can calculate it as follows:
Net force = Applied force - Frictional force
Net force = 175 N - 120 N
Net force = 55 N
Now, using Newton's second law of motion, we can determine the acceleration of the crate. Rearranging the equation, we have:
Net force = mass * acceleration
55 N = 50 kg * acceleration
Solving for acceleration:
acceleration = 55 N / 50 kg
acceleration = 1.1 m/s²
Since we know the initial velocity of the crate is zero (as it starts from rest), and we want to find the final velocity after 0.5 seconds, we can use the equation of motion:
final velocity = initial velocity + (acceleration * time)
Plugging in the values:
final velocity = 0 + (1.1 m/s² * 0.5 s)
final velocity = 0.55 m/s
Therefore, the crate's final velocity after 0.5 seconds is 0.55 m/s. This means that after being subjected to a 175 N force and experiencing 120 N of friction, the crate gains a velocity of 0.55 m/s in the direction of the applied force.
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In a game of pool, a 0. 4 kg cue ball is traveling at 0. 80 m/s when it hits a slower striped ball moving at 0. 38 m/s. After the collision, the striped ball moves off at 0. 62 m/s. What is the magnitude of the final velocity of the cue ball? Assume all pool balls have the same mass. 0. 20 m/s 0. 56 m/s 1. 0 m/s 1. 8 m/s.
When solving the problem of pool game and calculating the magnitude of the final velocity of the cue ball, the correct option is 0.56 m/s.
The following method: Use the principle of conservation of momentum, i.e. momentum before the collision is equal to the momentum after the collision, which is mathematically written as: [tex]$$mv_1+Mv_2=(m + M)v_3$$[/tex]
Where, m is the mass of the cue ball,
M is the mass of the striped ball,
v1 is the velocity of the cue ball before the collision,
v2 is the velocity of the striped ball before the collision, and
v3 is the velocity of the cue ball after the collision.
Using the above formula, we get the final velocity of the cue ball as:
[tex]$$v_3=frac {mv_1+Mv_2}{m+M}$$[/tex]
Plug in the given values, we get,
[tex]$$v_3=frac{0.4*0.80+0.4*0.38}{0.4+0.4}$$[/tex]
Solving for v3, we get [tex]$v_3=0.59$[/tex] m/s Therefore, the magnitude of the final velocity of the cue ball is 0.59 m/s.
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Assuming a constant density, the size of an object scales as its mass raised to what power?.
Assuming a constant density, the size of an object scales as its mass raised to the power of 1/3 (one-third).
The mass, density, and volume of an object are related by the equation:
ρ = m/Vwhere ρ is the density, m is the mass, and V is the volume.
We can write this equation as
V = m/ρThis equation can be used to find the relationship between the mass and volume of an object of constant density.
Assume that we have two objects of the same material with masses m1 and m2.
We can find the ratio of their volumes by taking the ratio of their masses and density as follows:
V1/V2 = m1/ρ / m2/ρV1/V2 = m1/m2V1/V2 = (m1/m2)^(1/3)
This shows that the ratio of the volumes of two objects with the same density is proportional to the cube root of the ratio of their masses.
This relationship can be expressed as:
V ∝ m^(1/3)
This relationship can also be expressed as the size of an object scales as its mass raised to the power of 1/3.
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