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sineoko [7]
2 years ago
10

Which of these is another name for Newton's

Physics
2 answers:
dezoksy [38]2 years ago
5 0
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
tino4ka555 [31]2 years ago
5 0
D. The law of inertia
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What is the critical angle of a light beam passing drim a medium(n=2) to a medium (n=1.5)
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Explanation:

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3 years ago
A boy is playing with a ball of mass 72g attached to a string. He is moving it at constant speed in a horizontal circle of radiu
irina [24]

Answer:

Explanation:

radius of circle r = 0.9 m.

(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .

( b )

Tension in string T = m ω²r

Putting the values

60 = .072 x ω² x 0.9

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angle made in 20 revolutions θ = 20 x 2π = 126.6 rad

time taken = θ / ω

= 126.6 / 30.4

= 4.16 s .

3 0
3 years ago
The thunderbolt bobsled team is training for Olympic Gold. During practice they start a run with a speed of 0.57 m/s, they compl
aleksandr82 [10.1K]
acceleration=\frac{\Delta\ velocity}{\Delta\ time}\\\\
v_{initial}=0,57m/s\\
distance=1360m\\ \Delta\ time=89,49seconds\\\\
v_{final}-v_{initial}=\frac{distance}{time}\\
v_{final}=\frac{distance}{time}+v_{initial}\\
v_{final}=\frac{1360}{89,49}+0,57\\\\v_{final}=15,77\frac{m}{s}\\\\
acceleration=\frac{15,77-0,57}{89,49}=0,17\frac{m}{s^2}\\\\ \boxed{acceleration=0,17\frac{m}{s^2}}
6 0
3 years ago
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor
Fofino [41]

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      Em_{f} = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

3 0
3 years ago
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