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3 years ago

For which group of people might brain-powered cars allow the greatest increase in mobility?

Physics

2 answers:

dolphi86 [110]3 years ago

8 0

Brain-powered cars?
The answer would be people who lack motor skills in their legs.

8090 [49]3 years ago

5 0

Answer:

People with physical disabilities

Explanation:

The brain-driven car allows an individual to drive a car without touching the steering wheel, only through sensors integrated with software installed on a notebook.

These cars promise to have a major impact on science as the system enables people with paralysis or other physical disabilities to have full control of items around them. One type of helmet, equipped with electroencephalographic sensors, has the function of capturing the driver's brainwaves, while a program, developed especially for the project, captures the results, turning them into commands for the car.

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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

$V_1=V_2=50912\ V$

The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

6 0

3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

7 0

1 year ago

The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti

deff fn [24]

Answer:

Explanation:

Given

side of square shape $a=5\ cm$

Electric flux $\phi =3\times 10^{-5}\ N.m^2/C$

Permittivity of free space $\epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}$

Flux is given by

$\phi =EA\cos \theta$

where E=electric field strength

A=area

$\theta$ =Angle between Electric field and area vector

$E=\frac{\phi }{A\cos (0)}$

$E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}$

$E=0.012\ N/C$

and Electric field by a uniformly charged sheet is given by

$E=\frac{\sigma }{2\epsilon_0}$

where $\sigma$ =charge density

$=\frac{\sigma }{\epsilon_0}$

$\sigma =0.012\times 8.85\times 10^{-12}$

$\sigma =2.12\times 10^{-13}\ C/m^2$

5 0

3 years ago

What is the lowest mass that an object can have and still be a star?

CaHeK987 [17]

Answer:

The lowest mass that an object can have to be considered a star is 0.08 solar masses.

Explanation:

A star is get when it reaches the necessary temperature in its core to nuclear reaction began.

A Nuclear reaction is the fusion of lighter elements into heavier elements.

In stars there is an equilibrium between two forces, the force of gravity in the inward direction due to its own mass and the radiation pressure in the upward direction as a consequence of the nuclear reaction in its core, which is known as hydrostatic equilibrium.

Therefore, the mass of the star must be enough to the force of gravity act in the inward direction, which leads to the increase in pressure, density and of course temperature in the core, allowing the nuclear reaction to begin.

Hence, the lowest mass that an object can have to be consider a star is 0.08 solar masses.

5 0

3 years ago

Help!!!

sattari [20]

Answer:

Mass is 152.95743

weight in ibs 3304.6914215069355 ibs

EXPLANATION:

1500 N = 152.95743 KG

152.95743 KG × 9.8 = 1498.982814

1 KG is 2.2 IBS

14o8.982814 is 3304.6914215069355

6 0

2 years ago