A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separation s could you produce in a typical atom by applying
s = 6.25 10⁻²² m
Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by
p = q s
s = p / q
s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹
s = 0.625 10⁻²¹ m
s = 6.25 10⁻²² m
We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.
A screw can be considered a type of
Fastener because a fastener is something that connects to objects and usually can come apart but can also be permanent
) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region
- z direction
To find the direction of the magnetic field, you take into account that the magnetic force over a charge, is given by the following cross product:
F_B: magnetic force
q: charge of the particle
v: velocity of the charge
B: magnetic field
In this case you have that the electron is moving along x-axis. You can consider this direction as the ^i direction. The electron experiences a magnetic deflection in the -y direction, that is, in the -^j direction.
By the cross products between unit vectors, you have that:
-^j = ^i X ^k
That is, the cross product between two vectors, one in the +x direction, and another one in the +z direction, generates a vector in the -y direction. However, it is necessary to take into account that the negative charge of the electron change the sign of the result of the cross product, which demands that the second vector is in the -z direction. That is:
-^i X -k^ = ^i X ^k = - ^j
Hence, the direction of the magnetic field is in the -z direction
How many joules of work are done on an object when a force of 10 N pushes it 5 m?A) 2 J B) 5 J C) 50 J D) 1 J E) 10 J
Force on the object = 10 N
distance of push = 5 m
Work done = ?
work done is equal to Force into displacement.
W = F . s
W = 10 x 5
W = 50 J
Work done by the object when 10 N force is applied is equal to 50 J
Hence, the correct answer is option C
The work done on an object when a force of 10 N pushes it 5 m is 50 Joules, calculated by multiplying the force and the displacement. So, the correct option is C.
The question is asking about work, which in physics is the result of a force causing a displacement. The formula for work is defined as the product of the force (in Newtons) and the displacement (in meters) the force causes. If a force of 10 N pushes an object a distance of 5 m, the work done is calculated by multiplying the force and the displacement (10 N * 5 m), yielding 50 Joules of work.
An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?
(a) The acceleration of the bird is . The negative sign indicated the opposite direction of motion. (b) The final speed is .
The acceleration can be computed from the velocities and time. The standard unit of acceleration is a meter per second square.
The acceleration is computed as:
Hence, the acceleration of the bird is . The negative sign indicated the opposite direction of motion.
The final speed as the given time can be computed from the first equation of motion. The first equation of motion gives the relation between final and initial speed, acceleration, and time.
A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.True False
A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.
The answer is false
A pressure antinode in a sound wave is indeed a region of high pressure, while a pressure node is a region of low pressure. These definitions hold true for all types of waves.
That's true. In terms of sound waves, a pressure antinode is a region of high pressure, while a pressure node is a region of low pressure. This is true for all types of waves, not only sound waves. In essence, a wave moves through a medium (in case of a sound wave, that medium is typically air) by creating areas of high and low pressure - the high pressure areas are called antinodes, and the low pressure areas are called nodes.