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prohojiy [21]
2 years ago
9

An astronaut is drifting away from his spaceship. how can he return, using an aerosol spray?

Physics
1 answer:
beks73 [17]2 years ago
6 0

The astronaut should shoot the aerosol spray in the opposite direction of where the astronaut is going to return back to the spaceship. This is in accordance to Newton’s Thrid Law, every action has an equal and opposite reaction.


Hope this helps you out!

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Why is the answer B?
djyliett [7]

Answer:

Explanation:

The center of mass lies on a line that joins position 4 of one start with position 4 of the other star.  The shortest distance between these two points will produce the largest velocity. You are using F = m v^2/R

Small R = large force.

Large Force = increased speed.

The masses don't have any effect on the outcome: they remain constant.

7 0
3 years ago
The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec
notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

<u>λ = 1.8 x 10⁻⁷ m = 180 nm</u>

5 0
3 years ago
A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its
sdas [7]
<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

           s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² -  u x 2 - 0.5 x g x 2²

           s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² -  u x 4 - 0.5 x g x 4²

           s₅ = u - 4.5 g    

That is

           u - 2.5 g = u - 4.5 g + 3

             2 g = 3

                g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

8 0
2 years ago
A net force of 345 N accelerates a boy on a sled at 3.2 m/s^2 . What is combined mass of the sled
Daniel [21]

Answer:

Mass, m = 26.54kg

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force
  • Fapp is the applied force
  • Fg is the force due to gravitation

<u>Given the following data;</u>

Net force, Fnet = 345

Acceleration, a = 3.2m/s²

<u>To find mass;</u>

Fnet = Fapp + Fg

Fnet = ma + mg

Fnet = m(a+g)

m = Fnet/(a+g)

We know that acceleration due to gravity, g = 9.8m/s²

Substituting into the equation, we have;

m = 345/(3.2 + 9.8)

m = 345/13

Mass, m = 26.54kg

6 0
2 years ago
You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
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