In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with radius 5.0 m. The ride moves at a constant speed and makes one complete circle in a time T=4.0s. What is the acceleration of the passengers? If the ride increases in speed so that T=3.0s, what is arad? (This question can be answered by using proportional reasoning, without much arithmetic.)
From the data given, the radius is 5.0m, and the time taken to complete one circle is 4.0secs
Since the motion is in a circular part, we can conclude that the total distance covered in this time is given as circumference of the circle.
which is expressed as
To determine the speed, we use the equation
The acceleration as required is expressed as
if the speed increase and it takes 3secs to complete one circle, the speed is
and the acceleration becomes
The acceleration of the passengers in the vertical circle carnival ride is 19.6 m/s^2. When the time taken to complete one circle is 3.0 s, the new acceleration is 26.13 m/s^2.
The acceleration of the passengers can be determined using the centripetal acceleration formula, which is given by a = v^2 / r.
In this case, the velocity v can be found by dividing the circumference of the circle (2πr) by the time taken to complete one circle (T). The radius r is given as 5.0 m. Plugging in the values, we have:
a = (v^2) / r = ((2πr / T)^2) / r = (4π^2r) / T^2 = (4π^2 * 5.0) / 16.0 = 19.6 m/s^2
To find the new acceleration when the time taken to complete one circle is 3.0 s, we can use the proportional reasoning to determine the relationship between the two accelerations. Since the time is inversely proportional to the acceleration, when T is 3.0 s, the new acceleration arad can be found using the equation:
What If? Fluoride ions (which have the same charge as an electron) are initially moving with the same speed as the electrons from part (a) through a different uniform electric field. The ions come to a stop in the same distance d. Let the mass of an ion be M and the mass of an electron be m. Find the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a). (Use the following as necessary: d, K, m, M, and e for the charge of the electron.)
E₁ / E₂ = M / m
Let the electric field be E₁ and E₂ for ions and electrons respectively .
Force on ions = E₁ e where e is charge on ions .
Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0
v² = u² - 2as
0 = u² - 2 x E₁ e d / M
u² = 2 x E₁ e d / M
Similarly for electrons
u² = 2 x E₂ e d / m
2 x E₁ e d / M = 2 x E₂ e d / m
E₁ / E₂ = M / m
The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.
The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:
Work = Change in kinetic energy
The change in kinetic energy can be calculated using the formula:
Change in kinetic energy = (1/2)Mv2 - (1/2)mv2
where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:
Ratio = (1/2)M/(1/2)m = M/m
So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.
A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).
time is 32 s and speed is 304.3 m/s
Height, h = 146 m
speed, u = 14 m/s
Angle, A = 43 degree
Let it hits the ground after time t.
Use second equation of motion
Time cannot be negative so the time is t = 32 s .
The vertical velocity at the time of strike is
v' = u sin A - g t
v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s
v'' = 14 cos 43 =10.3 m/s
The resultant velocity at the time of strike is
On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs account for 80 % . We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 61.0-kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. The skater is initially spinning at 70.0 rpm with his arms outstretched.Required: What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?
To find the final angular velocity when the skater pulls in his arms, we use the conservation of angular momentum.
To find the final angular velocity when the skater pulls in his arms, we can make use of the conservation of angular momentum. Initially, the skater's arms are outstretched, and the moment of inertia can be calculated using the parallel axis theorem. After the skater pulls in his arms, we can calculate the new moment of inertia using the same theorem. Equating the initial and final angular momentum values, we can solve for the final angular velocity.
The problem involves the concept of conservation of angular momentum. The skater's spinning speed will increase when they pull their arms in. For a precise value of the final velocity, a complex calculation taking into account body mass distribution is needed.
This question involves the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant as long as no external torques act on it. The total initial angular momentum of the skater spinning with outstretched arms is equal to his final angular momentum when he pulls his arms in.
Calculating the skater's initial and final angular momentum, you can then solve for his final velocity.
However, note that the calculation needs to take into account the skater's mass distribution. Specifically, we need to consider the percentage distributions for the arms/hands (13%), head (7%) and trunk/legs (80%), and integrate these over the skater's body.
This can result in a significantly complex calculation if done accurately, involving calculus level mathematics. However, using the qualitative knowledge that the skater's spinning speed will increase when they pull their arms in, it's reasonable to estimate, considering the mass distribution, the final velocity will be somewhere near 2 to 3 times the original rpm. But for an exact value, a detailed and complex calculation is needed.
Learn more about Conservation of Angular Momentum here:
Is it true or false if a force is a push or a pull on a object
Explanation: a force can be anything that effects an object, as long as the object moves
Explanation:Because anything that is being pushed of pulled is called force.
HOPE IT HELPED :)
A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that when placed in the field the sphere is in a static situation (all the forces on the sphere cancel). If the thread is horizontal, find the magnitude and direction of the electric field. The sphere has a mass of 0.018 kg and contains a charge of + 6.80 x 103 C. The tension in the thread is 6.57 x 10-2 N. Show your work and/or explain your reasoning. (20 pts)
direction is Horizontal
As we know that the string is horizontal here
so the tension force in the string is due to electrostatic force on it
now we will have
so here the force is tension force on it
now we have
direction is Horizontal
The magnitude of the electric field on the charged sphere in this scenario is approximately 1.17 x 10^-5 N/C. The direction of the electric field is horizontal, which is the same direction as the tension in the thread.
To start, we can use the equilibrium condition where the tension in the thread is equal to the force due to the electric field and gravity on the sphere. The formula to calculate the electric force is F = qE, and the gravitational force is F = mg, where F is the force, q is the electric charge, E is the electric field, m is the mass of the object, and g is the gravity constant.
Tension - electric force - gravitational force equals zero: T - F_electric - F_gravity = 0. We fill in the previous formulas: T - qE -mg = 0. This can be rearranged to E = (T + mg) / q.
In this case, the sphere's mass m is 0.018 kg, the tension T is 6.57 x 10^-2 N, and the sphere's charge q is 6.80 x 10^3 C, and we use g = 9.81 m/s². So, E = ((6.57 x 10^-2) + (0.018 * 9.81)) / 6.80 x 10^3.
This leads to an electric field magnitude of approximately 1.17 x 10^-5 N/C. The direction of the electric field is the same as the direction of the tension, which is horizontal due to the thread being horizontal.