Assemblage is an additive process where artists construct work by putting together objects and attaches them in some way. The correct option among all the options that are given in the question is the second option or option "b". This kind of artistry actually gives a three dimensional impression to the work that is done by the artist.
Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.
Answer:
Explanation:
Generally, length of vector means the magnitude of the vector.
So, given a vector
R = a•i + b•j + c•k
Then, it magnitude can be caused using
|R|= √(a²+b²+c²)
So, applying this to each of the vector given.
(a) 2i + 4j + 3k
The length is
L = √(2²+4²+3²)
L = √(4+16+9)
L = √29
L = 5.385 unit
(b) 5i − 2j + k
Note that k means 1k
The length is
L = √(5²+(-2)²+1²)
Note that, -×- = +
L = √(25+4+1)
L = √30
L = 5.477 unit
(c) 2i − k
Note that, since there is no component j implies that j component is 0
L = 2i + 0j - 1k
The length is
L = √(2²+0²+(-1)²)
L = √(4+0+1)
L = √5
L = 2.236 unit
(d) 5i
Same as above no is j-component and k-component
L = 5i + 0j + 0k
The length is
L = √(5²+0²+0²)
L = √(25+0+0)
L = √25
L = 5 unit
(e) 3i − 2j − k
The length is
L = √(3²+(-2)²+(-1)²)
L = √(9+4+1)
L = √14
L = 3.742 unit
(f) i + j + k
The length is
L = √(1²+1²+1²)
L = √(1+1+1)
L = √3
L = 1.7321 unit
Answer:
1.70 J
Explanation:
The heat dissipated is the difference in the kinetic energies.
This is given by
and 
are the initial and final velocities.
With <em>m</em> = 0.175 kg,
The negative sign appears because energy is lost.
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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