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Mariana [72]
4 years ago
15

Introduction This section describes the principles and concepts related to the experiment. It is used to help the person reading

your report to understand the information that serves as foundation for the experiment. It also includes a brief statement about the purpose of the lab. Example: Surface tension is the attractive force exerted upon the surface molecules of a liquid by the molecules beneath. Water has a high surface tension because there are strong attractive forces between water molecules. Adding surfactants, like soap, can reduce the surface tension of a liquid. The purpose of the experiment was to determine the effect of soap on the surface tension of water.
Physics
1 answer:
ArbitrLikvidat [17]4 years ago
6 0

Answer:

See explanation

Explanation:

It is a common observation that when water drops on a clean glass surface, it appears as little bags of water.

Surface tension makes water surface act like a stretched elastic skin bag.

Surface tension is caused by attractive forces between the molecules at the surface and the molecules in the bulk of a liquid. Remember that the forces of cohesion between water molecules are very strong.

When surfactants such as soap are added to water, the surface tension of water is decreased, hence the experiment.

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As a freely falling object speeds up, what is happening to its acceleration - does it increase, decrease, or stay the same? (a)
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Answer:

(a)When ignoring air resistance its accelerating increases steadily .

(b)When considering air resistance then its acceleration decreases this could either be uniformly or unevenly.

Hope this helped.

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3 years ago
The elbow is an example of a compound machine found in the human body. true or false
GenaCL600 [577]
The elbow is considered to be a lever, which is a simple machine. Other levers apart from the elbow are jaw, ankle, knee, etc.
This means that it is false that the <span>elbow is an example of a compound machine found in the human body.</span>
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4 years ago
Two forces,
serg [7]

First compute the resultant force F:

\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N

\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N

\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N

Then use Newton's second law to determine the acceleration vector \mathbf a for the particle:

\mathbf F=m\mathbf a

(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a

\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}

Let \mathbf x(t) and \mathbf v(t) denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so \mathbf v(0)=0. Then the particle's velocity vector at <em>t</em> = 10.4 s is

\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du

\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}

\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}

If you don't know calculus, then just use the formula,

v_f=v_i+at

So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component

v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}

(b) Compute the angle \theta for \mathbf v(10.4\,\mathrm s):

\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ

so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.

(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

Then using the fundamental theorem of calculus again, we have

\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du

where \mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m is the particle's initial position. So we get

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}

\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m

So over the first 10.4 s, the particle is displaced by the vector

\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

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3 years ago
Which form of energy body will possess when it is kept on the top of Hill
Oduvanchick [21]

Potential energy would be the answer


:)

3 0
3 years ago
Read 2 more answers
Someone help please i’ll give brainliest
Fantom [35]
Directly proportional to volume, according to charles’s law
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