4. Which are characteristics of an atom? (Choose all that apply.)

Explain a situation in which you can accelerate even though your speed doesn’t change.

Serga [27]

"Acceleration" does NOT mean speeding up. It also doesn't mean
slowing down. Acceleration means ANY change in the speed
OR DIRECTION of motion.

The only kind of motion that's NOT accelerated is motion at a steady
speed AND in a straight line.

Even when your speed is steady, you're accelerating if your direction
is changing.

A few examples:
(no speeds are changing):

-- driving on a curved road, or turning a corner
-- going around a curve on a skateboard, a bike, or a Segway
-- running on a quarter-mile track
-- an Indy car cruising a practice lap around the track
-- water spinning, getting ready to go down the drain
-- any point on the blade of a fan
-- the little ball going around the inside of a Roulette wheel
-- the Moon in its orbit around the Earth
-- the Earth in its orbit around the sun

4 0

3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

g Two radiation modes (one at the center frequency lIo and the other at lIO+?lI) are excited with 1000 photons each. Determine t

Schach [20]

Answer:

a) P=0.25x10^-7

b) R=B*N2*E

c) N=1.33x10^9 photons

Explanation:

a) the spontaneous emission rate is equal to:

1/tsp=1/3 ms

the stimulated emission rate is equal to:

pst=(N*C*o(v))/V

where

o(v)=((λ^2*A)/(8*π*u^2))g(v)

g(v)=2/(π*deltav)

o(v)=(λ^2)/(4*π*tp*deltav)

Replacing values:

o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2

the probability is equal to:

P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7

b) the rate of decay is equal to:

R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system

c) the number of photons is equal to:

N=(1/tsp)*(V/C*o)

Replacing:

N=100/(3*3x10^10*8.3x10^-19)

N=1.33x10^9 photons

7 0

3 years ago

Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0

lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0

3 years ago

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is_____

Bond [772]

When you are talking about calculating the kinetic energy of an object, the formula is as stated:

Ek =1/2mv²

Where m is mass and v is velocity. Sub all those numbers in and you'll get 25J of kinetic energy.

3 0

3 years ago