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3 years ago

A technique in which people use machines to learn how to control their bodies is known as __________.

Physics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:

The technique in which people use machines to learn how to control their bodies is known as D, Biofeedback.

Explanation:

Biofeedback is a variety of different machines that help people learn how to control their bodies depending on their specific needs, varying from things like scalp sensors, electrocardiographs, electromyographs and more.

Dafna11 [192]3 years ago

4 0

Answer:

biofeedback

Explanation:

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A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

5 0

1 year ago

(a) Calculate the acceleration due to gravity on the surface of the Sun.

Ira Lisetskai [31]

<h2>a)Acceleration due to gravity on the surface of the Sun is 274.21 m/s²</h2><h2>b) Factor of increase in weight is 27.95</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

Here we need to find acceleration due to gravity of Sun,

G = 6.67259 x 10⁻¹¹ N m²/kg²

Mass of sun, M = 1.989 × 10³⁰ kg

Radius of sun, r = 6.957 x 10⁸ m

Substituting,

$g=\frac{6.67259\times 10^{-11}\times 1.989\times 10^{30}}{(6.957\times 10^8)^2}\\\\g=274.21m/s^2$

Acceleration due to gravity on the surface of the Sun = 274.21 m/s²

b) Acceleration due to gravity in earth = 9.81 m/s²

Ratio of gravity = 274.21/9.81 = 27.95

Weight = mg

Factor of increase in weight = 27.95

8 0

3 years ago

How many degrees are in each quadrant <br> A: 90°<br> B: 30°<br> C: 180°<br> D: 360°

Airida [17]

In one quadrant there are 90 degrees.

8 0

3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

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3 years ago

an automobile has a tire pressure of 325 kpa when the temperature is 10°C of the temperature rises to 50°C what is the new press

Finger [1]

Old temperature = 283 K.
New temp = 323 K.

(323/283) x (325 kPa) = 371 kPa.

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3 years ago