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Cloud [144]
3 years ago
10

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2.500×10^3 Newtons while Magnus pulls the bus a distance of 2.312.31 meters, how much work does the tension force do on Magnus? The rope is perfectly horizontal during the pull.
Physics
1 answer:
ozzi3 years ago
7 0

Answer:

The total work that the rope does to Mangnus is - 5780 Jules.

Explanation:

By definition, the work is defined as:

W=F.d

Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.  

Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.  

For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement:

W=-2500 N * 2.312 m

W=-5780 Nm = -5780 J

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Setler [38]

Answer:

a) 1.2\times 10^2\ N

Explanation:

t = Time taken

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v = Final velocity

a = Acceleration

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The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

3 0
3 years ago
If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity ch
Slav-nsk [51]
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8 0
3 years ago
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harkovskaia [24]

Answer:

m = 3 kg

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Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

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7 0
3 years ago
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topjm [15]

Answer:

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Explanation:

Given that,

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Using formula of time period

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4.74}

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Hence, The period of oscillation is 1.33 sec.

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Alja [10]
Nickel has a happy amount of 28 electrons.
8 0
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