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dusya [7]
3 years ago
14

What is the average velocity of the particle from rest to 15 seconds? A. 1.1 meters/second B. 1.2 meters/second C. 1.3 meters/se

cond D. 1.4 meters/second

Physics
1 answer:
Over [174]3 years ago
8 0

Answer: D. 1.4 meters/second

Explanation:

To calculate average Velocity, the proportion of the total Displacement or distance traveled to the total time taken during that Displacement is calculated.

Total Displacement from rest(that is 0) to 15 seconds (21, on the graph)

(0, 21)

Total time taken = 15 seconds

Average Velocity = (total Displacement / total time taken)

Average Velocity = (21 - 0) / (15 - 0)

Average Velocity = 21 / 15

Average Velocity = 1.4 meters per second

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Which statement best reflects the approach of Gestalt psychology?
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A person pushes a refrigerator across a horizontal floor. The mass of the refrigerator is 110 kg, the coefficient of static fric
Law Incorporation [45]

Answer:

Explanation:

mass of refrigerator, m = 110 kg

coefficient of static friction, μs = 0.85

coefficient of kinetic friction, μk = 0.59

(a) the minimum force required to just start the motion in refrigerator

F = μs x mg

F = 0.85 x 110 x 9.8

F = 916.3 N

(b) The force required to move the refrigerator with constant speed

F' = μk x mg

F' = 0.59 x 110 x 9.8

F' = 636.02 N

(c) Let a be the acceleration.

Net force = Applied force - friction force

F net = 950 - 636.02

F net = 313.98 N

a = F net / mass

a = 313.98 / 110

a = 2.85 m/s²

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
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