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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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What is the equivalent resistance between the points A and B of the network?

Dominik [7]

Explanation:

First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.

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3 years ago

Why is Darwin criticized for his theory? and who were they

kondaur [170]

A few people criticized Darwin for his theory, such as the left-leaning biologists Stephen Jay Gould and Richard Lewontin, who fear the political implications of Darwinian theory. They fear that evolutionary theory, even when bolstered by modern genetics, and molecular biology, does not make reality probable enough.

4 0

2 years ago

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A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

6 0

3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

wariber [46]

<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $\frac{11.7}{18}$

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

5 0

3 years ago

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As you know, the oceans cover most of the earth, and contain a huge amount of resources. Protecting these resources is one reaso

matrenka [14]

A is the answer your looking for

8 0

3 years ago

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