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3 years ago

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An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 times 10^6 m and a mass of 7.35 times

10^22 kg. An experiment is planned where a projectile needs to be launched straight up from the surface. What must be the minimum initial speed of the projectile so it will reach a height of 2.55 times 10^6 m above this satellite's surface? (G = 6.67 times 10^-11 N middot m^2/kg^2)

1 answer:

Eduardwww [97]3 years ago

6 0

Answer:

aaaaaaaaaaaaaaaaaaaaaahh sorry

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The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

$\vartriangle y =\frac{m \lambda L}{d}$

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

We simply replace the values in that equation :

$\vartriangle r= m \lambda =2.\ 460\ nm$

$\vartriangle r= 920\ x\ 10^{-9} m$

The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.

3 0

3 years ago

Ejercicio 1: Un cuerpo gira en un círculo de 8cm de diámetro con una rapidez constante

Lera25 [3.4K]

Answer:

Exercise 1;

The centripetal acceleration is approximately 94.52 m/s²

Explanation:

1) The given parameters are;

The diameter of the circle = 8 cm = 0.08 m

The radius of the circle = Diameter/2 = 0.08/2 = 0.04 m

The speed of motion = 7 km/h = 1.944444 m/s

The centripetal acceleration = v²/r = 1.944444²/0.04 ≈ 94.52 m/s²

The centripetal acceleration ≈ 94.52 m/s²

8 0

3 years ago

A reconnaissance plane flies 605 km away from

kolezko [41]

Answer:

$v_{avg} =$ 355 m/s

Explanation:

Distance = 605 km

Initial speed = $v_{i}$ = 284 m/s

Final velocity = $v_{f}$ = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

$2as = v_{f}^{2}+v_{i}^{2}$

Then using first equation of motion, we find time

$v_{f} = v_{i}+at$

Then using the formula of average velocity, we find average velocity

$v_{avg}=\frac{total-distance}{total-time}$

Second method is very simple

$v_{avg}=\frac{v_{f}+v_{i} }{2}$

$v_{avg}=\frac{426+284}{2}$

$v_{avg} =$ 355 m/s

8 0

3 years ago

3 Draw energy transfer diagrams

timofeeve [1]

Answer:

The outline of the energy transfer are;

a) Kinetic energy → Clockwork spring → Potential energy

b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

Please find attached the drawings of the energy transfer created with MS Visio

Explanation:

The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred

a) The energy transfer diagram for the winding up a clockwork car is given as follows;

Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;

Kinetic energy → Clockwork spring → Potential energy

b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;

Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries

Therefore, we have;

Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

6 0

2 years ago

Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

Norma-Jean [14]

Answer:

Part a)

$\lambda = 300 m$

Part b)

$E = 2.7 N/C$

Part c)

$I = 9.68 \times 10^{-3} W/m^2$

$P = 3.22 \times 10^{-11} N/m^2$

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

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

$\lambda = \frac{3\times 10^8}{1\times 10^6}$

$\lambda = 300 m$

Part b)

As we know the relation between electric field and magnetic field

$E = Bc$

$E = (9 \times 10^{-9})(3\times 10^8)$

$E = 2.7 N/C$

Part c)

Intensity of wave is given as

$I = \frac{1}{2}\epsilon_0E^2c$

$I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)$

$I = 9.68 \times 10^{-3} W/m^2$

Pressure is defined as ratio of intensity and speed

$P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}$

$P = 3.22 \times 10^{-11} N/m^2$

6 0

3 years ago