3 years ago

Hey guys! Am I right? Thanks!

Physics

1 answer:

worty [1.4K]3 years ago

5 0

Answer:

the correct answer is reduce friction

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A series of pulses of amplitude 0.25 m are sent down a string that is attached to a post at one end. The pulses are reflected at

Klio2033 [76]

Answer:

Explanation:

since the string is rigidly attached to the post, the reflected and incident pulses are of the same amplitude but different polarities. At the point where the two pulses meet, the amplitude will be the addition of that of the incident and the reflected pulses. i.e Amplitude= 0.25 + -0.25=0m

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4 years ago

A tank contains 0.568 mol of molecular nitrogen (N2). Determine the mass (in grams) of nitrogen that must be removed from the ta

Marizza181 [45]

Answer:

$\Delta M = 7.812 gm$

Explanation:

we know that for ideal gag we have

pV =nRT

Since volume, gas constant R and T are constant, so we have

$\frac{p}{n] = constant$

$\frac{44.7}{0.568} =\frac{22.8}{n}$

n = 0.289 mole

hence mass removed $\Delta M =( 0.568 - 0.289)*( molecular\ weight\ of \ nitrogen)$

$\Delta M = (0.568 - 0.289) *28 gm$

$\Delta M = 7.812 gm$

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4 years ago

The droplets on a glass of ice water are a result of water vapor releasing energy and changing to water droplets.

olya-2409 [2.1K]

Answer:

B. condensation

Explanation:

hope this helps

4 0

3 years ago

Read 2 more answers

When all parts of a circuit are composed of conducting materials, the circuit is said to be

Korolek [52]

Closed is the correct answer :)

7 0

4 years ago

A force in the x direction acts on a particle moving also in the x direction, producing a potential energy U(x)=Ax4 where A=0.63

poizon [28]

Answer:

The magnitude of the force is 1.29*10^-3N in the positive x direction

Explanation:

In order to calculate the magnitude and direction of the force, you take into account that the force is the space derivative of the potential enrgy, as follow:

$F(x)=-\frac{dU(x)}{dx}$ (1)

where:

$U(x)=Ax^4\\\\A=0.0630\frac{J}{m^4}$

You replace the expression for U into the equation (1) and solve for F:

$F(x)=-\frac{d}{dx}[Ax^4]=-4Ax^3$ (2)

The force on the particle, for x = -0.080m is:

$F=-4(0.630\frac{J}{m^4})(-0.0800m)^3=1.29*10^{-3}N$

The magnitude of the force is 1.29*10^-3N in the positive x direction

7 0

3 years ago