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LekaFEV [45]
3 years ago
5

His the bottom, or lowest point, of a wave.

Physics
2 answers:
Gnom [1K]3 years ago
5 0

A trough is the lowest point in a wave.

Olin [163]3 years ago
5 0

Answer:

The lowest part is the trough.

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An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move
kupik [55]

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4.7ft

Explanation:

Pls see attached file

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3 years ago
The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?
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Read 2 more answers
A. l1 =20 A and l2 =24 A
Masteriza [31]

Answer:

when u find out pls lmk! i have the same question and I've been stuck for a while lol

4 0
3 years ago
If we decrease the amount of force, and kelp all other factors the same, what will happen to the amount of work?
Artist 52 [7]

Answer:

Distance will decrease and work will decrease:

F = m a      Newton's Second Law

a = F / m       decreasing force will decrease acceleration

S = 1/2 a t^2 = 1/2 (F / m) t^2     distance traveled will decrease as force        decreases

W = F * S       work will decrease as both force and distance decrease

8 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
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