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exis [7]
4 years ago
13

A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25

cm. if the pulling force exerted on the mass was 80.N, determine the spring constant of the spring.
Physics
1 answer:
solniwko [45]4 years ago
3 0

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

k=\frac {80}{0.25}=320N/m

Therefore, the spring constant is equal to 320 N/m

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Answer:

I=2.766\ kg.m^2

Explanation:

We have:

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weight of the wheel, w_w=280\ N

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height of descend, h=2.5\ m

(a)

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I=\frac{1}{2} m.r^2

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I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

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4 years ago
A 19kg block is being pulled with a constant horizontal force of 95 Newton’s while also experiencing a constant friction force o
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A

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3 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
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Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

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Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

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0.295 = 3.18 - 4.905t²

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t = √2.885/4.905

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v = 3.18/(0.5882 × cos45)

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