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Alekssandra [29.7K]
3 years ago
13

An object O is acted upon by a 6 N force and a 5 N force. The angle between these two forces is 60°. Draw a scale diagram showin

g the forces acting on object O and determine the resultant force.​
Physics
1 answer:
FrozenT [24]3 years ago
6 0

Answer:

Resultant force = \sqrt{91}N

Explanation:

Rf = \sqrt{f1²+f2²+2.f1.f2.cos60}

Rf = \sqrt{6²+5²+2.6.5.1/2}

Rf = \sqrt{36+25+30}

Rf = \sqrt{91}

doesn't look right though...

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8 0
3 years ago
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7 0
3 years ago
How is work affected when an object is lifted straight up instead of using a ramp? Work will increase.
kifflom [539]
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3 0
3 years ago
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

Hence, this is the required solution.

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3 years ago
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katovenus [111]

Hi there


The correct answer is : C

Wave


I hope that's help:0

6 0
4 years ago
Read 2 more answers
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