What is a Super Massive Black Hole?

7. A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0

Illusion [34]

Answer:

The distance between first-order and second-order bright fringes is 12.66mm.

Explanation:

The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

$\Lambda x = L\frac{\lambda}{d}$ (1)

Where $\Lambda x$ is the distance between two adjacent maxima, L is the distance of the screen from the slits, $\lambda$ is the wavelength and d is the separation between the slits.

The values for this particular case are:

$L = 2.0m$

$\lambda = 633nm$

$d = 0.100mm$

Notice that is necessary to express L and $\lambda$ in units of milimeters.

$L = 2.0m \cdot \frac{1000mm}{1m}$ ⇒ $2000mm$

$\lambda = 633nm \cdot \frac{1mm}{1x10^{6}nm}$ ⇒ $6.33x10^{-4}mm$

Finally, equation 1 can be used:

$\Lambda x = (2000mm)\frac{(6.33x10^{-4}mm)}{(0.100mm)}$

$\Lambda x = 12.66mm$

Hence, the distance between first-order and second-order bright fringes is 12.66mm.

8 0

3 years ago

Which term below best describes the forces on an object with a a net force of zero?

dybincka [34]

Answer:

B. Balanced Forces

Explanation:

The net force is defined as the sum of all the forces acting on an object. Therefore, if the forces are balanced, they will counteract each other, causing the net force to be zero, then the object will continue at rest or moving with constant velocity.

4 0

4 years ago

If a person uses about 4 calories to fuel a minute of walking, and it takes about 20 minutes to walk a mile, approximately how m

Lemur [1.5K]

Answer:

43.75 miles must a person walk to utilize the energy in (“burn”) a pound of fat.

Explanation:

3,500 calories are present in 1 pound of the fat.

Thus, given that:

4 calories are burnt in 1 minute of walking.

So,

1 calories are burnt in 1/4 minute of walking.

Or,

1 calories are burnt in 0.25 minute of walking.

Thus,

3500 calories are burnt in 0.25*3500 minutes of walking

Minutes of walking needed to burn 3500 calories = 875 minutes.

Also, given that:

20 minutes of walking covers 1 mile.

1 minute of walking covers 1/20 mile.

So,

875 minutes of walking covers (1/20)*875 mile.

43.75 miles must a person walk to utilize the energy in (“burn”) a pound of fat.

6 0

3 years ago

An electric motor is used to run an elevator. The total mass of the elevator car and passenger is 1600 kg. The elevator moves up

Masja [62]

Answer:

Approximately $2.7 \times 10^{4}\; {\rm W}$ , assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$

Explanation:

The weight of the elevator is:

$\begin{aligned}& \text{weight} \\ =\; & m\, g \\ =\; & 1600\; {\rm kg} \times 9.81\; {\rm m \cdot s^{-2}} \\ \approx\; & 15700\; {\rm N}\end{aligned}$ .

Since the speed of the elevator is constant, the acceleration of this elevator would be $0$ .

By Newton's Second Law of Motion, the net force on the elevator (proportional to acceleration) would also be $0\!$ . All external forces on the elevator need to be balanced in every direction.

The only two vertical forces on the elevator are:

the weight of the elevator (downward gravitational pull from the earth,) and
the upward pull from the motor.

These two forces need to balance one another. Since the weight of the elevator is approximately $15700\; {\rm N}$ , the upward pull of the motor would be $15700\; {\rm N}\!$ . in magnitude.

The direction of this upward pull is the same as the direction of the motion of this elevator. Thus, the work that the motor did on the elevator would be positive:

$\begin{aligned}& \text{work} \\ =\; & F\, s \\ \approx\; & 15700\; {\rm N} \times 15.0\; {\rm m} \\ \approx\; & 2.35 \times 10^{5}\; {\rm J}\end{aligned}$ .

Since the velocity of the elevator is constant, instantaneous power output of the motor would be equal to the average power of the motor:

$\begin{aligned}& \text{power} \\ =\; & \frac{\text{work}}{t} \\ \approx\; & \frac{2.35 \times 10^{5}\; {\rm J}}{8.82\; {\rm s}} \\ \approx\; & 2.7 \times 10^{4}\end{aligned}$ .

4 0

3 years ago

Which type of wave forms at the boundary between air and water when you

azamat

D. Transverse wave

Hope it's correct and helps uh.

6 0

3 years ago

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