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3 years ago

If the small piston has an area of .004 square

Physics

1 answer:

denpristay [2]3 years ago

8 0

The magnitude of the pressure : P = 56,250 N/m

<h3>Further explanation</h3>

Given

Area of 0.004 m²

Force 225 N

Required

the magnitude of the pressure

Solution

Pressure = force exerted per unit of surface area(perpendicular)

Can be formulated :

P = F/A

Input the value :

P = 225 N / 0.004 m

P = 56,250 N/m

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A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals.

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time should you wait between pushes is 2.83 sec.

the question is incomplete, full statement is-

A 24 kg child sits on a 2.0-m-long rope swing. You are going to give the child a small, brief push at regular intervals. If you want to increase the amplitude of her motion as quickly as possible, how much time should you wait between pushes?

<h3>What is Amplitude?</h3>

In physics, amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.

regular interval - at similarly spaced intervals: having the same interval of time between occurrences From 4 a.m. to midnight, the buses operate at regular intervals. The boards are positioned at regular intervals, with an equal amount of space between each.

The length of swing, l = 2.1 m

The time between the pushes is nothing but the Time period

and is given by the formula,

$T = 2\pi ( \frac{l}{g} )^{\frac{1}{2} }$

= 2 * 3.14 ( 2.0/ 9.8 ) ^ (1/2)

= 2.83 sec

to learn more about Amplitude go to - brainly.com/question/3613222

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1 year ago

Which instrument is used to measure liquid volume?

Alecsey [184]

Volumetric cylinders and volumetric flasks

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3 years ago

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A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

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3 years ago

Here is a force diagram of an object in water. The weight of the object is 15N and the buoyancy force is 17N. Will the object fl

lawyer [7]

Answer:

Object will float.

Explanation:

Total force on the body = Weight of body + Buoyancy force on body.

Weight of body = 15 N downwards = 15 N

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Total force on body = 15 - 17 = -2 N = 2 N upwards

So, the body will float.

Object will float.

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3 years ago

"In a Young’s double-slit experiment, the separation between slits is d and the screen is a distance D from the slits. D is much

san4es73 [151]

Answer:

The number of bright fringes per unit width on the screen is, $x=\dfrac{\lambda D}{d}$

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and $\lambda$ is the wavelength of the light. Let x is the number of bright fringes per unit width on the screen is given by :

$x=\dfrac{n\lambda D}{d}$

$\lambda$ is the wavelength

n is the order

If n = 1,

$x=\dfrac{\lambda D}{d}$

So, the the number of bright fringes per unit width on the screen is $\dfrac{\lambda D}{d}$ . Hence, the correct option is (B).

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2 years ago