All categories

3 years ago

What is the purpose of the article physicists test telepathy in a cheat proof setting?

1 answer:

7 0

I'm guessing that the purpose of the article is to inform the
general public of the work that these Physicists did when they
attempted to detect the existence or non-existence of telepathy
in a controlled, scientifically acceptable setting.

The REASON the article was ever presented in a non-professional
publication was most likely in deference to the general public's interest
in telepathy, UFOs, aliens, flat Earth, Astrology, and all other forms of
weird garbage. In other words, its publication was motivated more by
the desire to sell magazines than by any inclination to impart any real,
useful information to the general public.

You might be interested in

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

p_2 = ?

t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

7 0

4 years ago

A pair of adult nitwits sit balanced on a teeter totter type device. One of them, whose mass is 54.2 kg, is 1.20 m from the poin

andreev551 [17]

Answer:

The mass of the second person is 28.91 kg

Explanation:

Given;

mass of the first adult, m₁ = 54.2 kg

distance of the first adult from the point of balance, x = 1.20 m

mass of the second adult, = m₂

distance of the second adult from the point of balance, y = 2.25 m

Taking moment about the point of balance, we will have

m₁x = m₂y

54.2 x 1.2 = 2.25y

2.25y = 65.04

y = 65.04/2.25

y = 28.91 kg

Therefore, the mass of the second person is 28.91 kg

3 0

4 years ago

Read 2 more answers

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

4 years ago

A ball of mass 4.5 kg moving with speed of 2.2 m/s in the +x-direction hits a wall and bounces back with the same speed in the x

Levart [38]

Answer:

The change of the momentum of the ball is $-19.8\, \frac{mkg}{s}$

Explanation:

We should find $\varDelta\overrightarrow{p}=\overrightarrow{p_{f}}-\overrightarrow{p_{i}}$ (1)with $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum. Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (1):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (2)

It's important to note that momentum and velocity are vectors and direction matters, so if +x direction is the direction towards the wall and the -x direction away the wall $\overrightarrow{v_{i}}=+2.2\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-2.2\, \frac{m}{s}$ so (2) becomes: