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3 years ago

6. A balloon inflated with helium is able to float toward the ceiling because

Physics

1 answer:

Nostrana [21]3 years ago

5 0

Answer:

b) the density of the balloon is less than the density of the atmosphere.

Explanation:

We must take into account that density is defined as the relationship between mass and volume.

The helium density is 0,1785 [kg/m3]

The air density is 1.2 [kg/m3]

Therefore a substance with lower density (helium) will try to rise on substances with heavier density (air).

An example of this is when we mix water with oil, the water is heavier than the oil, that is, the density of the water is greater than the density of the oil, therefore the oil will be located on the surface of the water and not at the bottom.

Another example is that of a balloon inflated with air is common, this balloon doesn't rise because the air density outside the balloon equals the air density inside the balloon, the balloon will fall to the ground since it contains the weight of the air contains the weight of the material of the balloon.

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which of the following will not increase the strength of an electromagnet made by wrapping a wire around an iron nail

serg [7]

what is the following?

8 0

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 t = 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 t = 10.4 s has x-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 x axis.

(c) We can find the velocity at any time t 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).

3 0

3 years ago

Einstein’s general theory of relativity made or allowed us to make predictions about the outcome of several experiments that had

PtichkaEL [24]

Answer:

1. bending of light in gravitational fields.

2. effect of gravitational redshift.

3. perihelion precission of mecury.

Explanation:

1 bending of light in gravitational fields, we can think of it like this:

by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.

2. effects of gravitational redshift:

this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.

3. perihelion precission of mecury:

according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.

3 0

4 years ago

How would you get your friend to change his/her mind from using PED's?

kvasek [131]

Answer:

Watch funny videos and start eating a lot of candy

6 0

3 years ago

Question 2

Nostrana [21]

Answer:

in pretty sure the last one

8 0

4 years ago