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arlik [135]
3 years ago
11

An object moves along a straight line path from P to Q under the action of a force (4 3 3 ) N. I j k − + If the coordinates of P

and Q, in metres, are (3, 3, -1) and (2, -1, 4) respectively, then the work done by the force is
Physics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

1 Nm

Explanation:

Given;

Force = F = (4, 3, 3)N

Position 1 = P = (3, 3, -1)m

Position 2 = Q = (2, -1, 4)m

The object moves along a straight line path from P to Q, therefore, the distance vector (d) is given by;

d = Q - P

d = (3, 3, -1) - (2, -1, 4)

d = (1, 4, -5)m

Now the work done (W) by the force (F) to move through the distance (d) is the dot product of the two vectors: F and d. i.e

W = F . d

For clarity, let's write vectors F and d in vector unit notation as follows;

F = 4 i + 3 j + 3 k

d = 1 i + 4 j - 5k

Therefore,

W = (4 i + 3 j + 3 k ) . (1 i + 4 j - 5k)

W = (4 + 12 - 15)

W = 1

Therefore, the workdone by the force is 1 Nm

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Information gathered from observing a plant grow 3 cm over a two week period is<br> called:
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Answer:

Data

- Information gathered from observing a plant grow 3 cm over a two week period is called data.
8 0
3 years ago
The Milky Way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to Hubble
barxatty [35]

Answer: SBb

On 1930 the astronomer Edwin Hubble classified the galaxies based on their visual appearance into elliptical, spiral and irregular, being the first two classes the most frequent.  

So, according to this classification, the Milky Way is a barred spiral galaxy (SBb in Hubble's notation system) because it has a central bar-shaped structure of bright stars that spans from one side of the galaxy to the other. In addition, its spiral arms seem to emerge from the end of this "bar".

Scientifics considered this, after measuring the the disk and central bulge region of the galaxy, and the conclusion is the Milky Way fulfills these conditions, because is a galaxy that orbits on its same axis and with this rotation its arms are twisted in opposite directions around the mentioned axis.

7 0
3 years ago
Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when t
yawa3891 [41]

Answer:

a) x=2*10^{9} m and t=8.35 s

b) t = γt', so it is 8.35 s.

Explanation:

a) The equation of Lorentz transformations is given by:

x=\gamma(x'+ut')  

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

  • x' = 0
  • t' = 5.00 s
  • u =0.800 c, c is the speed of light 3*10⁸ m/s

\gamma=\frac{1}{\sqrt{1-(u/c)^{2}}}

\gamma=\frac{1}{\sqrt{1-(0.800c/c)^{2}}}

\gamma=\frac{1}{\sqrt{1-(0.800)^{2}}}

\gamma=1.67

x=1.67(0+0.800c*5.00)

x=2*10^{9} m

Now, to find t we apply the same analysis:

t=\gamma(t'+\frac{ux'}{c^{2}})                        

but as x'=0 we just have:

t=\gamma(t')

t=1.67*5.00=8.35 s

b) Here, Mavis reads 5 s on her watch and Stanley measured the events at a time affected by the Lorentz factor, in other words t = γt', if we see it is the same a) part. So the time interval will be equal to 8.35 s.

I hope it helps you!

5 0
3 years ago
Read 2 more answers
A basketball of mass 0.608 kg is dropped from rest from a height of 1.37 m. It rebounds to a height of 0.626 m.
kenny6666 [7]

Answer:

a)|\Delta E|=4.58\: J  

b)F=61.90\: N

Explanation:

a)

We can use conservation of energy between these heights.

\Delta E=mgh_{2}-mgh_{1}=mg(h_{2}-h_{1})  

\Delta E=0.608*9.81(0.6026-1.37)

Therefore, the lost energy is:

|\Delta E|=4.58\: J  

b)

The force acting along the distance create a work, these work is equal to the potential energy.

W=\Delta E

F*d=mgh

Let's solve it for F.

F=\frac{mgh}{d}

F=\frac{0.608*9.81*1.37}{0.132}

Therefore, the force is:

F=61.90\: N

I hope is helps you!

6 0
2 years ago
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bixtya [17]

Answer:

There are four types of orbitals that you should be familiar with s, p, d and f (sharp, principle, diffuse and fundamental). Within each shell of an atom there are some combinations of orbitals

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2 years ago
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