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Inessa [10]
3 years ago
6

A 5-kg object slides down a frictionless surface inclined at an angle of 30º from the horizontal. The total distance moved by th

e object along the plane is 10 meters. The work done on the object by the normal force of the surface is?
Physics
1 answer:
erma4kov [3.2K]3 years ago
7 0

Answer:

Workdone =0

Explanation:

The body is inclined at θ=30°,     s=10m,  m=5kg

Workdone by the object = FsCosθ

but the surface is a friction-less surface, this means coefficient of friction =0

F=μR

F=0*R

F=0

W=FScosθ

W=0*SCosθ

W=0

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an astronaut on an eva has wandered dangerously far away from the shuttle. she has also exhausted all the fuel in her jet pack.
V125BC [204]

The conservation of the momentum allows to find the result of how the astronaut can return to the spacecraft is:

  • Throwing the thruster away from the ship.

The momentum is defined as the product of the mass and the velocity of the body, for isolated systems the momentum  is conserved. If we define the system as consisting of the astronaut and the evo propellant, this system is isolated and the internal forces become zero. Let's find the moment in two moments.

Initial instant. Astronaut and thrust together.

        p₀ = 0

Final moment. The astronaut now the thruster in the opposite direction of the ship.

       m_f = m v + M v '

where m is propellant mass and M the astronaut mass.

As the moment is preserved.

       0 = m v + M v ’

      v ’= - \frac{m}{M} \ v  

We can see that the astronaut's speed is in the opposite direction to the propeller, that is, in the direction of the ship.

The magnitude of the velocity is given by the relationship between the masses.

In conclusion, using the conservation of the momentun we can find the result of how the astronaut can return to the ship is:

  • Throwing the thruster away from the ship.

Learn more here:  brainly.com/question/14798485

5 0
2 years ago
In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi
Ronch [10]

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

6 0
3 years ago
A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating
Rainbow [258]

Answer:

41.4496148484\ m/s

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

\sigma = Surface charge density = 5.9\times 10^{-8}\ C/m^2

\Delta x = 0.57-0.26

q = Charge = 6.9\times 10^{-9}\ C

m = Mass of object = 8.3\times 10^{-9}\ kg

Electric field due to a sheet is given by

E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m

Electric field is given by

E=\dfrac{V}{d}

Voltage is given by

V=E\Delta x

Kinetic energy is given by

K=qV

\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s

The initial speed of the object is 41.4496148484\ m/s

7 0
3 years ago
Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m
maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

6 0
3 years ago
Which bright solar feature is shown in the picture above?
Ghella [55]

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called <em>Prominence</em>.

<em>Photosphere</em> is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

<em>Corona</em> is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

6 0
3 years ago
Read 2 more answers
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