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Zielflug [23.3K]
2 years ago
5

A football goalkeeper moves across her goal in a straight line. Her motion is shown on the following graph of horizontal positio

n xxx vs. time ttt. Graph of x (in meters) vs. t (in seconds). The y-intercept is at (0,0), then x increases linearly from 0 m to 2 m over 4 s, stays constant at 2 m from 4 to 8 seconds, and then decreases linearly from 2m to 0 m between 8 s and 12 s. Graph of x (in meters) vs. t (in seconds). The y-intercept is at (0,0), then x increases linearly from 0 m to 2 m over 4 s, stays constant at 2 m from 4 to 8 seconds, and then decreases linearly from 2m to 0 m between 8 s and 12 s. What is the average speed of the goalkeeper between the times t=4\text{ s}t=4 st, equals, 4, start text, space, s, end text and t=12\text{ s}t=12 st, equals, 12, start text, space, s, end text? Choose 1 answer: Choose 1 answer: (Choice A) A -0.25\,\dfrac{\text m}{\text s}−0.25 s m ​ minus, 0, point, 25, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction (Choice B) B 0.50\,\dfrac{\text m}{\text s}0.50 s m ​ 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction (Choice C) C -0.50\,\dfrac{\text m}{\text s}−0.50 s m ​ minus, 0, point, 50, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction (Choice D) D 0.25\,\dfrac{\text m}{\text s}0.25 s m ​
Physics
1 answer:
sergejj [24]2 years ago
5 0

Answer:

0.25 m/s

Explanation:

Acceleration cannot be negative.

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A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

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3 years ago
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E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??
luda_lava [24]

Explanation:

  • Mass(m)= 20kg
  • Acceleration (a)= 5m/s²
  • Force(F)= ?

We know that,

  • F=ma
  • F=20×5
  • F=100N

Hence, the needed force is 100N.

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1 year ago
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True or false The momentum of a 100kg object traveling at 3 m/s is 300 N
melomori [17]

Answer:

True

Explanation:

Momentum of an object can be defined as the product of its mass and velocity at which it is travelling. With that in mind, momentum = 3*100=300(kg⋅m/s).

One thing to note is the units mentioned. The SI unit of momentum is kg * m/s as it is the product of mass(kilograms) and velocity(meter per second) and not Newton.

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3 years ago
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3 years ago
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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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