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bixtya [17]
2 years ago
8

First, record some data for this comparison in the table below.

Physics
1 answer:
kondor19780726 [428]2 years ago
5 0

Answer:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).

Explanation:

Record your measured values of displacement and velocity for times t = 8.0 seconds and t = 10.0 seconds in the columns below.

Next, use the measured displacement and velocity values at t = 7.0 seconds and t = 9.0 seconds to interpolate the values of displacement and velocity at t = 8.0 seconds.

Use the following formula to interpolate and extrapolate. Remember, x and y here represent values on the x and y axes of the graph. The x values will really be time and the y values will be either displacement (x) or velocity (vx).

This is the answer

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Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
Mars2501 [29]

Answer:

hello your question has some missing parts below is the complete question

and the missing diagram

The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.

answer : 1316.2 hertz

Explanation:

The frequency that produce the maximum sound intensity can be calculated using the relation below

dsin ∅ = n <em>A</em>

where <em>A = </em>dsin ∅ / n  when n = 1 . d = 0.800

<em>A</em> = 0.800 * ( 1 / 3.162 )

<em>A</em> = 0.253 m

speed of sound = 333 m/s

frequency = speed /<em> A</em>

<em>=   </em>333 / 0.253 =  1316.2 hertz

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2 years ago
A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma
Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

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A race car driver must average 200km/hr for four laps to qualify for a race. Because of engine trouble, the car averages only 17
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The average speed would have to be 260 km/hr due to the driver originally going 30 km/hr too slow the first two laps
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2 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
Zanzabum

(a)  3.3\cdot 10^{-6} Pa

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

p=\frac{I}{c}

where

I is the intensity of the wave

c is the speed of light

In this problem,

I=1000 W/m^2

and substituting c=3\cdot 10^8 m/s, we find the radiation pressure

p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa

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A=6.65\cdot 10^{-29}m^2

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N

And then, since we know the mass of the atom

m=5.01\cdot 10^{-27}kg

we can find the acceleration, by using Newton's second law:

a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2

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A metal bar moves through a magnetic field. the induced charges on the bar are
Dmitry [639]
I would say its a positive cgarge
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