Answer:
Acceleration = 2.35 m/
Speed = 8.67 m/s
Explanation:
The coefficient of friction , u =0.3
The angle of incline = 30°
The two forces acting on block are weight and friction.
weight along the incline = mg cos60° = 
= 0.5 mg
Friction along incline = umg cos30° = mg 
Friction along incline = 0.26 mg
Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg
Acceleration = 
= 0.24 g = 2.35 m/
The height of incline = 8 m
Length of the inclined edge = 16 m
v= 8.67 m/s
Answer:
A. New space missions show that Pluto is much larger than originally thought.
Explanation:
The new definition of a planet that was adopted in 2006, defined planet as an object that orbits the sun, with sufficient mass to be round, not a satellite of another object, and has removed debris and small objects from the area around its orbit.
This new definition of a planet that was adopted in 2006, classified Pluto as "dwarf planet", because Pluto meets planetary criteria except that it has not cleared debris from its orbital neighborhood.
However, new Horizons spacecraft flew by Pluto in 2015, revealed that Pluto is much larger than originally thought
Therefore, the correct option is "A"
A. New space missions show that Pluto is much larger than originally thought.
Work done is given by the change in kinetic energy of an object
- The kinetic energy of the shovel, the shrub, and in Robert's movement were changed, therefore, work is done in the given processes,
Reason:
Work is done when the total energy of object is affected by the application of force on the object over a distance
Therefore;
- In option <em>A</em>, pushing the shovel into ground (to dig out the dirt) the requires the application of a force (push) over a distance, (into and out of the ground) therefore work is done
- In option <em>B</em>, picking the shrub up gives it gravitational potential energy, therefore, work is done
- In option <em>C</em>, carrying the shrub to the hole does visible work
- In option <em>D</em>, holding the shrub while lowering it into the hole does work by preventing the shrub from falling randomly
Therefore, <u>work is done in the given processes</u>
Learn more about work-energy theorem here:
brainly.com/question/10063455
Answer:
F=(-4.8*10^22,0,0) N
Explanation:
<u>Given :</u>
We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°
Solution :
We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.
The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next
F=|F|cosФp (1)
Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet
F=|F|cosФp
=-4.8*10^22 N*p
<em>As this force is in one direction, we could get its vector as next </em>
F=(-4.8*10^22,0,0) N
F=(0,-4.8*10^22,0) N
F=(0,0-4.8*10^22) N
The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.
Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx= 
replacing ;
Δx = 2t ; we have:
2t =
Given that thickness t = 700 nm
Then
2× 700 = --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness 
∴ 
Thus, the smallest thickness t of the air film = 140 nm
