A charge of -0.4 µC is located at the origin; a charge of 0.53 µC is located at x = 0.2 m, y = 0; a third charge Q is located at x = 0.32 m, y = 0. The force on the 0.53 µC charge is 4.3 N, directed in the positive x direction.

1. Determine the charge

2. With this configuration of three charges, where, along the x direction, is the electric field zero? xf=

3. x2=

Explain why it is difficult to make an X-ray laser.

What is the fastest transverse wave that can be sent along a given wire? For safety reasons, the maximum tensile stress to which this wire should be subjected is 3.56 × 108 N/m2. The density of the wire is 8300 kg/m3. Note that your answer does not depend on the diameter of the wire.

A 0.180-kg wooden rod is 1.60 m long and pivots at one end. It is held horizontally and then released.

Part A : What is the angular acceleration of the rod after it is released? Express your answer to three significant figures and include appropriate units.

Part B: What is the linear acceleration of a spot on the rod that is 0.560 m from the axis of rotation? Express your answer to three significant figures and include appropriate units.

Part C: At what location along the rod should a die be placed so that the die just begins to separate from the rod as it falls? Express your answer to three significant figures and include appropriate units.

At what radius along the perpendicular bisector of a wire of length 0.29 m carrying 30 nC of charge does the assumption of cylindrical symmetry in applying Gauss's law give you an answer whose error exceeds 5%?

Question1:playground merry-go-round has mass of 75kg and a radius of 3 m. A 25kg child sits on the merry-go-round 2/3's of way out from center.What's the moment of inertia of the merry-go-round/child system?

Question2:On the playground merry-go round in the previous question,one child pushes it CCW with a force of 20 N, while the 25 kg child drags her feet through a hole 2/3 of the way out from the center,causing a CW force of 17 N. What's angular acceleration of the merry-go-round?

Radiation from the head is a major source of heat loss from the human body. Model a head as a 21-cm-diameter, 20-cm-tall cylinder with a flat top.

If the body's surface temperature is 36?C , what is the net rate of heat loss on a chilly 9?C  day? All skin, regardless of color, is effectively black in the infrared where the radiation occurs, so use an emissivity of 0.95.

I have worked it through three times with other written solutions however i cannot arrive at the correct answer. A final answer as well as a solution would be appreciated! Thanks!

A- How much work is required to accelerate a proton from rest up to a speed of 0.995c?

W = ? J

B- What would be the momentum of this proton?

P = ? kg⋅m/s

What is the basic, fundamental meaning of the Continuity equation? What are some examples and the importance of this rule?

GOAL Apply the more general definition of torque. PROBLEM (a) A man applies a force of F = 3.00 102 N at an angle of 60.0° to the door of Figure (a), 2.00 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door? STRATEGY Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force. SOLUTION (A) Compute the torque due to the applied force exerted at 60.0°. Substitute into the general torque equation. τF = rFsin θ = (2.00 m)(3.00 ✕ 102 N) sin 60.0° = (2.00 m)(2.60 ✕ 102 N) = 5.20 102 N · m (B) Calculate the force exerted by the wedge on the other side of the door. Set the sum of the torques equal to zero. τhinge + τwedge + τF = 0 The hinge force provides no torque because it acts at the axis (r = 0). The wedge force acts at an angle of −90.0°, opposite the upward 260 N component. 0 + Fwedge(1.50 m) sin (−90.0°) + 5.20 ✕ 102 N · m = 0 Fwedge = 347 N LEARN MORE REMARKS Notice that the angle from the position vector to the wedge force is −90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque. QUESTION To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob? closer to the hinge closer to the doorknob PRACTICE IT Use the worked example above to help you solve this problem. (a) A man applies a force of F = 3.00 102 N at an angle of 60.0° to a door, x = 2.10 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. N · m (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door? N EXERCISE HINTS: GETTING STARTED | I'M STUCK! A man ties one end of a strong rope 7.74 m long to the bumper of his truck, 0.521 m from the ground, and the other end to a vertical tree trunk at a height of 3.62 m. He uses the truck to create a tension of 7.50 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point. N · m

a) If 8.00 hp are required to drive a 1800-kg automobile at 54.0 km/h on a level road, what is the total retarding force due to friction, air resistance, and so on?

b) What power is necessary to drive the car at 54.0 km/h up a 10.0% grade (a hill rising 10.0 m vertically in 100.0 m horizontally)?

c) What power is necessary to drive the car at 54.0 km/h down a 1.00 % grade?

d) Down what percent grade would the car coast at 54.0 km/h ?

A 0.0038 kg leaf falls 16.8 m to the ground, where it lands with a speed of 12.2 m/s. What was the average force of air resistance on the leaf while it was falling?

Problem 1: In the figure to the right, m1=20.0kg and α=53.1o . The coefficient of kinetic friction between the block and the incline is µk=0.40. a) What must be the mass m2 of the hanging block if it is to descend 12.0 m in the first 2.00 s after the system is released from rest? b) For the µk in the problem, what mass m2 will provide constant velocity of m1 down the incline? c) Suppose µs is 0.7, what is the maximum mass m2 that will still cause no movement?

For physics discussion post. Please TYPE the response as I struggle with reading written information.

• Explain Boyle's law, Charles law, and the Ideal Gas equation.
• Explain Kinetic Molecular Theory of Ideal Gas.
• In your own words, explain the First Law of Thermodynamics and give examples.
• In your own words, explain adiabatic process and isothermal process, and give examples.

Vector A = (-2, 2.4) and vector B = (4, 2.3). Find the magnitude of the component of A perpendicular to the direction of B