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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
10 mph i jus got it right on the test
Answer:
The focal length is 
The radius of curvature is 
Explanation:
From the question we are told that
The magnification of the mirror is 
The distance of the person from the mirror(the object distance ) is 
The negative sign shows that it is been placed in front of the mirror
Generally the magnification of the mirror is mathematically represented as
=> 
=> 
Generally from the lens equation we have that
=> 
=>
Generally the radius of curvature is mathematically represented as
=> 
=>
Answer:
q = 7.542 x 10⁻⁷ C = 754.2 nC
Explanation:
The Coulomb's Law gives the magnitude of the force of attraction or repulsion between two charges:
F = kq₁q₂/r²
where,
F = Force of attraction or repulsion = 0.2 N
k = Coulomb's Constant = 9 x 10⁹ N m²/C²
r = distance between charges = 16 cm = 0.16 m
q₁ = magnitude of 1st charge
q₂ = magnitude of 2nd charge
Since, both charges are said to be equal here.
q₁ = q₂ = q
Therefore,
0.2 N = (9 x 10⁹ N m²/C²)q²/(0.16 m)²
(0.2 N)(0.16 m)²/(9 x 10⁹ N m²/C²) = q²
q = √(5.88 x 10⁻¹³ C²)
<u>q = 7.542 x 10⁻⁷ C = 754.2 nC</u>
Answer:
The work done on the athlete is approximately 2.09 J
Explanation:
From the definition of the work done by a variable force:
and substituting with the function of our problem:
