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2 years ago

What is the normal force in the image below?

Physics

1 answer:

saveliy_v [14]2 years ago

3 0

Answer:

980.7N

Explanation:

F = m * g

F = 100 * 9.807m/ $s^{2}$

F = 980.7N

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What is the difference between torque and the moment of a force

STatiana [176]

Answer:

Torque Of a Force: If The Force has tendency or Bends The Body about Longitudinal axis of the Body it is Torque. Moment Of a Force :If Force has Tendency to or Rotates the Body about Transverse asis the Body It is Moment .

Explanation:

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2 years ago

Does the diagram show a spring or neap tide?How do you know?

vampirchik [111]

It shows a neap tide

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3 years ago

A parallel-plate capacitor is disconnected from a battery, and the plates are pulled a small distance farther apart. Does Q incr

kolezko [41]

Answer:

Q stay the same

Explanation:

Charging of capacitor is done by battery . If battery is disconnected , charging will stop . There will not be any discharging as plates are separate . So pulling the plates apart will not affect the charge lying on the capacitor . It will decrease its capacity and increase its potential , keeping its charge constant.

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3 years ago

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ( $a_{r}$ ), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ( $T$ ), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)