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3 years ago

Any body moving with simple harmonic motion is being acted on by a force that is:__________.

Physics

1 answer:

Stella [2.4K]3 years ago

6 0

Answer:

Explanation:

Because this oscillations occur when the restoring force is directly proportional to displacement, given as

F=-kx

Where k= force constant

X= displacement

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A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge

navik [9.2K]

The solution would be like this for this specific problem:

F=−k∗x∗q∗Q/(+)F−≈k∗x∗q∗Q/R3[(1−3/2*x2/R3]
F=−k∗x∗q∗/QR3
F=ma
−k∗q∗Q/R3*x=ma
−k∗x=m∗a
a==ω2x
ω=(k/m)1/2

ω=(kqQ/R3)1/2

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

5 0

3 years ago

Two straight wires carry current of 5A opposite direction separated by a distance of 30cm. What is the magnitude of magnetic fie

jek_recluse [69]

Answer:

The magnitude of magnetic field of both wires is β= 13. 33 $x10^{-6}$ T

Explanation:

$p=15cm*\frac{1m}{100cm}=0.15m$

$I=5A$

$u_{o} = 4*\pi x10^{-7}$

Using the right hand rule field magnetic as the current go in opposite direction the field in the point exactly in the middle have the same direction both so

$\beta _{t} =\beta _{1}+\beta _{2}$

They have the same direction and the P point is the middle as the same current field can be find:

$\beta _{t} =\beta _{1}*2$

$\beta = \frac{u_{o}*I }{2*\pi *p} \\\beta =\frac{4*\pi x10^{-7}*5A }{2*\pi *\frac{30}{2} } \\\beta =\frac{4x10^{-7}*5 }{0.30} \\\beta =6.6x10^{-6}$ T

$\beta _{t} =6.6x10^{-6}*2$ =13.33 $x10^{-6}$ T

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3 years ago

Fossil fuels like gasoline or coal are examples of

disa [49]

Answer:

Nonrenewable Resources

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3 years ago

STATE THE LAWS OF CONSERVATION OF MOMENTUM

dalvyx [7]

$\Huge \mid \underline {\mathcal {{{\color{purple}{Answer...}}}}} \mid$

When two bodies collide with each other in the absence of an external force, then the total final momentum of the bodies is equal to their total initial momentum.

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3 years ago

A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horiz

Rufina [12.5K]

Answer:

0.88752 kgm²

0.02236 Nm

Explanation:

m = Mass of ball = 1.2 kg

L = Length of rod = 0.86 m

$\theta$ = Angle = 90°

Rotational inertia is given by

$I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2$

The rotational inertia is 0.88752 kgm²

Torque is given by

$\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm$

The torque is 0.02236 Nm

5 0

3 years ago