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Lorico [155]
3 years ago
11

What is the most commonly used reference point

Physics
2 answers:
snow_tiger [21]3 years ago
7 0

Answer:

El punto de referencia más usado es el NORTE. Los puntos de referencia sirven para orientarnos, de esta manera podemos saber nuestra ubicación y luego desplazarnos. En este caso tenemos que el punto cardinal NORTE es el más usado, se tiene como referencia universal para lograr ubicarnos.

Explanation:

gracias

Marina86 [1]3 years ago
7 0
The most common combinations of reference points are the population average with the historical average, and the population average with the individual expectation.
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Microphone Electricity needs
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Sound—energy<span> we can hear—travels only so far before it soaks away into the world around us. Until electrical </span>microphones<span>were invented in the late 19th century, there was no satisfactory way to send </span>sounds<span> to other places. You could shout, but that carried your words only a little further. You couldn't shout in New York City and make yourself heard in London. And you couldn't speak in 1715 and have someone listen to what you said a hundred years later! Remarkably, such things are possible today: by converting sound energy into electricity and information we can store, microphones make it possible to send the sounds of our voices, our music, and the noises in our world to other places and other times. How do microphones work? Let's take a closer look!</span>
3 0
3 years ago
A skater is spinning at 15.0 rad/s with a rotational inertia of 8.0 kg m? If she changes her rotational inertia to 5.0 kg mº, wh
mixas84 [53]

Answer:Idrk man

Explanation:

3 0
3 years ago
A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling
IrinaVladis [17]

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

-u as it is moving in opposite direction.

mc = 1200 kg, uc = 31.1 m/s

ms = 2830 kg, us = - u = ?

Using conservation of momentum

mc × uc + ms × us = 0

1200 × 31.1 - 2830 × u = 0

u = 13.18 m/s

4 0
4 years ago
If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = \frac{1}{2}kx^2   ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = \frac{1}{2}k\times0.021^2

or

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = \frac{1}{2}\times13605.44\times (0.062)^2

or

U = 26.149 J

Also,

Work done = change in energy

or

W = 26.149 - 3.0 = 23.149 J

4 0
3 years ago
A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

v = 7934.2 m/s

5 0
3 years ago
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