150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wag
1 answer:
Answer:
the friction force on this box is closest to 45.9 N.
Explanation:
given data
Weight of the box W = 150 N
accelerating uniformly = 3.00 m/s²
coefficient of kinetic friction = 0.400
coefficient of static friction = 0.600
solution
we know box does not move relative to the wagon
we get here friction force that is express as
friction force = mass × acceleration ..............1
here mass = weight ÷ g
mass = = 15.3 kg
put value in equation 1 we get
friction force = 15.3 kg × 3
friction force = 45.9 N
So, the friction force on this box is closest to 45.9 N.
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Answer:
Explanation:
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but modified with only one charge showing in the numerator of the expression.
The correct answer is <span>3) 
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In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
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In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
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Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 × m
separation L = 0.5 mm = 0.5 × m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 × m
λ = 2042 nm