Explain what would happen if we began Charge-to-Mass Ratio of Electrons experiment with a dot on the CRT (cathode-ray tube) screen instead of a line?

You are to fire a cannon and are trying to hit a target 300 meters due east of your current position. Your cannon fires projectiles with a mass of 2 kg and with an initial speed of 60 m/s. Gravity is pulling the projectile downward at an acceleration of 9.8 m/s2. There is a wind blowing from the north pushing the projectile southward with a force of 3 N. What is the angle θ up from the ground you must aim the cannon, and what is the angle φ the direction you must turn the cannon to the north? There is a large ridge between the cannon and the target, so your angle of inclination θ must be greater than 45 degrees. Use the acceleration vector given by the wind and gravity, the initial velocity, and the initial position (the origin), to find a position function for the projectile. This will depend on t, θ, and φ. Once you have this function, solve for t, θ, and φ.

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A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200- kg puck has a speed of 1.00 m/s at an angle of θ = 53.0° to the positive x axis. (a) Determine the velocity of the 0.300-kg puck after the collision. (b) Find the fraction of kinetic energy lost in the collision.

and please explain

Explain the relationship between GAST and UT1? What is the relationship between UT1 and UTC?

Consider an aluminum annular disk with an outer radius 63.2 mm and inner radius 7.9 mm. The mass of the disk is 464 grams (HINT: the parameters of the disk are the same as in the previous pre-lab).

The disk is allowed to rotate on a frictionless table with the rotation axis at its center. The disk has a small pulley rigidly mounted at the top concentrically. The pulley's radius is 12.1 mm, and the mass of the pulley is negligible. A string is wrapped around the pulley, and a hanging mass of 19.8 g is tied at the other end of the string. When the mass falls under gravity, it causes the aluminum annular disk to rotate. Ignore the string's mass, and assume that the string's motion is frictionless.

What is the angular speed of the aluminum disk when the mass has fallen 14.5 cm?
ω = _____ rad/s

How long does it take for the mass to reach this point?

t= _____s

Calculate the length of the de broglie wave of an electron and a proton moving with a kinetic energy of 1keV. For what values of kinetic energy will their wavelength be equal to 0.1 nm? m_e = 9.1.10 ³¹kg, e = 1.602.10^-19C, mp = 1.627.10^-27 kg

P1.- A high diver leaves the end of a 5.0 m high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine: (a) her initial velocity Vo (b) the maximum height reached, (c) the velocity Vf with which she enters the water.

P2.- A rollercoaster moves 200 ft horizontally and then rises 135 ft at an angle of 30 degrees above the horizontal. It next travels 135 ft at an angle of 40 degrees downward. What is its displacement from its starting point?

P3.- A model rocket is launched straight upward with an initial speed of 50 m/s. It accelerates with a constant upward acceleration of 2 m/s2 until its engines stop at an altitude of 150 m.

a) What is the maximum height reached by the rocket?

b) How long after liftoff does the rocket reaches its maximum height?

c) How long is the rocket in the air?

what does the H2 vector in this environment when a magnetic field in the air environment with relative magnetic permeability 1 has a value of H1= 3 ax -4ay +5az and enters the ferromagnetic environment on the xy plane and has a relative magnetic permeability 80 ?

Assessment

Friction

Friction resists motion. If an object is stationary, friction tries to keep it from beginning to move. If an object is moving, friction slows it down and tries to stop it.

In all cases all you need to do is add an extra arrow (vector) to your free-body diagram. This new arrow always points opposite the direction of motion. When you use Newton's law to sum the forces, there will be one more term in the equation.

The magnitude of this new arrow/term is always given by Ff = μ FN where

μ is the "coefficient of friction", a number (theat you generally look up on a table) that tells how hard it is to slide two objects in contact.FN is the "Normal Force", or how hard the surface pushes up on the object. Generally you find the Normal force by summing all the forces in the y-direction and solving for FN. Most often however, if there aren't any forces acting in the y-direction other than gravity and the normal force, then for horizontal surfaces, FN = mg. Therefore, Ff = μmgfor inclined plane surfaces, FN = mg cos θ. Therefore, Ff = μmg cos θ

Question 1 (1 point)

Match the following formulas about calculating friction:

Question 1 options:

123

Always works. FN can be found by summing all the y-dir forces, recognizing that the acceleration in the y-dir is (probably) zero, and solve for FN

123

Works whenever the surface is an inclined plane and there are no y-direction forces except for gravity and the normal force

123

Works whenever the surface is horizontal and there are no y-direction forces except for gravity and the normal force

1.

Ff = μmg

2.

Ff = μmg cos θ

3.

Ff = μ FN

Question 6 (1 point)

A 3 kg box sits on a ramp of 6 degrees where the coefficient of friction is 0.2. A 24 N force pulls the box uphill. Find the acceleration.

Your Answer:

Question 6 options:

Answer

Question 7 (1 point)

A 2 kg box sits on a ramp where the coefficient of friction is 0.4. Find the angle that will cause the box to slide downhill at constant velocity.

Hints:

constant velocity means a = 0 sum the forces (downhill pull and friction) and solve for θsin θ / cos θ = tan θtake the arctan

Your Answer:

Question 7 options:

Answer

Question 8 (1 point)

A 2 kg box sits on a ramp of 14 degrees where the coefficient of friction is 0.4. A string runs uphill over a pulley and back down to a hanging mass of 7 kg. Assuming the box on the ramp is pulled uphill by the weight of the hanging mass, find the acceleration.

Your Answer:

Question 8 options:

Answer

Question 9 (1 point)

A 2 kg box sits on a ramp of 14 degrees where the coefficient of friction is 0.2. A string runs uphill over a pulley and back down to a hanging mass of 9 kg. Assuming the box on the ramp is pulled uphill by the weight of the hanging mass, find the acceleration.

Your Answer:

Question 9 options:

Answer

Submit Assessment0 of 9 questions saved

1. A 5.40 g coin is placed 17.0 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.810 and μk = 0.460.

Part A: What is the maximum angular velocity with which the turntable can spin without the coin sliding? Express your answer with the appropriate units.

2. A 800 g ball moves in a vertical circle on a 1.09 m -long string. If the speed at the top is 4.30 m/s , then the speed at the bottom will be 7.82 m/s .

Part A: What is the ball's weight? Express your answer with the appropriate units.

Part B: What is the tension in the string when the ball is at the top? Express your answer with the appropriate units.

Part C: What is the tension in the string when the ball is at the bottom? Express your answer with the appropriate units.

1. Advanced skiers turn by sliding the backs of their skis across the snow. Since the fronts of their skis don’t move much, the skis end up pointed in a new direction.

a. The amount of sideways force that a skier must exert on the skis to slide them sideways is proportional to how hard the skis press down on the snow beneath them. Why?

b. Less skilled skiers sometimes turn without unweighting—they push their skis sideways so hard that the skis slide anyway. This technique is exhausting. Why does it require so much work?

c. Why is wax used on skis? Why wouldn’t WD-40 work as well as the wax? What is in WD-40?

A Kaon K⁺ breaks down into two particles. From amongst the following three particles:

π⁰ (neutral pion)

η⁰ (eta meson)

μ⁺ (anti-muon, identical to a muon except for its electric charge)

which one(s) can be formed as the products of the described break-down? (the other product is not known).

a. π⁰

b. η⁰

c. μ⁺

d. η⁰, π⁰

e. μ⁺, π⁰

f. μ⁺, η⁰

g. all three of them

h. none of them

Please reason your answer based on the various laws of conservation, and explain in precise terms why each one the three particles can or cannot be the product of the breakdown.

Thank you!