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STatiana [176]
3 years ago
9

A force of 75 N is applied to a spring, causing it to stretch 0.3 m. What is the spring constant of the spring?

Physics
1 answer:
muminat3 years ago
3 0
The answer is D. 250 N/m <span />
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<h2>Answer::</h2>

Humans (biosphere) built a dam out of rock materials (geosphere). Water in the lake (hydrosphere) seeps into the cliff walls behind the dam, becoming groundwater (geosphere), or evaporating into the air (atmosphere).','.

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Paying off your debts will most likely
g100num [7]
I think the answer is d
4 0
3 years ago
A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w
Annette [7]

Answer:

66 rpm

Explanation:

The period of oscillation is given by

T=2\pi \sqrt{\frac {m}{k}}

\frac {k}{m}=\frac {4\pi^{2}}{T^{2}} where  T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force=m\omega^{2} L

\omega=\sqrt{\frac {k\triangle L}{mL}} where \triangle L is the elongation, L is original length, \omega is the angular velocity

Substituting the equation of \frac {k}{m} into the above we obtain

\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}

\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s

6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm

6 0
4 years ago
Did i draw the tangent line correctly?​
Jet001 [13]

Answer:

yes i belevie so

Explanation:

8 0
3 years ago
A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg,
pickupchik [31]

Answer:

317.22

Explanation:

Given

Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s

You 69.7kg, cw 1.01m/s, at r

Poodle 20.2 kg, cw 1.01/2 m/s, at r/2

Mutt 17.7 kg, 3r/4

You

Relative

ω = v/r

= 1.01/1.93

= 0.522

Actual

ω = 0.945 - 0.522

= 0.42

I = mr^2

= 69.7*1.93^2

= 259.6

L = Iω

= 259.6*0.42

= 109.4

Poodle

Relative

ω = (1.01/2)/(1.93/2)

= 0.5233

Actual

ω = 0.945- 0.5233

= 0.4217

I = m(r/2)^2

= 20.2*(1.93/2)^2

= 18.81

L = Iω

= 18.81*0.4217

= 7.93

Mutt

Actual

ω = 0.945

I = m(3r/4)^2

= 17.7(3*1.93/4)^2

= 37.08

L = Iω

= 37.08*0.945

= 35.04

Disk

I = mr^2/2

= 93.1(1.93)^2/2

= 173.39

L = Iω

= 173.39*0.945

= 163.85

Total

L = 109.4+ 7.93+ 36.04+ 163.85

= 317.22 kg m^2/s

4 0
4 years ago
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