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aliya0001 [1]
3 years ago
10

The message refers to which of the following?

Physics
1 answer:
Oliga [24]3 years ago
7 0

The message is the information being communicated from one place to another.

It used to be called the "intelligence".  But as time went on, it became
harder to ignore the obvious fact that that was going too far, and the
label was changed to the more IQ-neutral "message".    

You might be interested in
Two 2.0 g plastic buttons each with + 40 nC of charge are placed on a frictionless surface 2.0 cm (measured between centers) on
EleoNora [17]

Answer:

a. There are three potential energy interaction. b. 2.16 m/s c. 2.16 m/s d. 0 m/s

Explanation:

a. There are three potential energy interaction.

Let the charges be q₁ = +40 nC, q₂ = +250 nC and q₃ = + 40 nC and the distances between them be q₁ and q₂ is r, the distance between q₂ and q₃ is r  and the distance between q₁ and q₃ is  r₁ = 2r respectively. So, the potential energies are

U₁ = kq₁q₂/r, U₂ = kq₁q₃/2r and U₃ = kq₂q₃/r

U = U₁ + U₂ + U₃ = kq₁q₂/r +  kq₁q₃/2r + kq₂q₃/r (q₁ = q₃ = q and q₂ = Q)

U = kqQ/r +  kq²/2r + kqQ/r = qk/r(2Q + q/2)

b. To calculate the final speed of the left 2.0 g button, the potential energy = kinetic energy change of the particle.

ΔU = -ΔK

0 - qk(2Q + q/2)/r = -(1/2mv² - 0). Since the final potential at infinity equals zero and the initial kinetic energy is zero.

So qk(2Q + q/2)/r = -1/2mv²

v = √[2qk(2Q + q/2)/mr] where m = 2.0 g r = 2.0 cm

substituting the values for the variables,

v = √[2 × 40 × 10⁻⁹ × 9 × 10⁹(2 × 250 × 10⁻⁹ + 40 × 10⁻⁹/2)/2 × 10⁻³ × 2 × 10⁻²]

v = √[360(500 × 10⁻⁹ + 20 × 10⁻⁹)/2 × 10⁻⁵]

v = √[720(520 × 10⁻⁹)/4 × 10⁻⁵] = 2.16 m/s

c. The final speed of the right 2.0 g button is also 2.16 m/s since we have the same potential energy in the system

d.

Since the net force on the 5.0 g mass is zero due to the mutual repulsion of the charges on the two 2.0 g masses, its acceleration a = 0. Since it starts from rests u = 0, its velocity v = u + at.

Hence,

v = u + at = 0 + 0t = 0 m/s

8 0
4 years ago
A uniform disk a uniform hoop and a uniform sphere are released at the same time at the top of an inclined ramp. They all roll w
lara [203]

Answer:

D. The sphere the disk and the hoop

Explanation:

This is because the sphere has inertial of

2/5mR²

Disk 1/2mR²

Hope mR²

So these are moment of inertial which is resistance or opposition to rotation so since the sphere has a smaller moment to inertial it will move faster and reach the ground first then the disk then the hoop in that order

6 0
3 years ago
When placed 1.18 m apart, the force each exerts on the other is 11.2 N and is repulsive. What is the charge on each
aleksandr82 [10.1K]

Answer:

q=41.62\ \mu C

Explanation:

Given that,

Force between two objects, F = 11.2 N

Distance between objects, d = 1.18 m

We need to find the charge on each objects. The force between charges is as follows :

F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C

So, the charge on each sphere is 41.62\ \mu C.

7 0
3 years ago
a solid object is found to weigh 4.784.78n in air. when it is weighed while fully immersed in water, its apparent weight is 2.48
ololo11 [35]

A solid object is found to weigh 4.784.78n in air. when it is weighed while fully immersed in water, its apparent weight is 2.482.48n. 983 is the density of the object.

The substance's density is defined as its mass per unit of volume (volumetric mass density or specific mass). Although the Latin letter D may also be used, the symbol for density that is most usually used is (the lower case Greek letter rho). where V is the volume, is the density, and m is the mass. Weight per unit volume is a common informal definition of density, however this is incorrect scientifically; the actual term is specific weight. The US oil and gas industry serves as one illustration of this. A pure substance's mass concentration in numbers is equal to its density. To make density comparisons between different systems of units easier, it is occasionally replaced by the dimensionless quantity "relative density" or "specific gravity," which is the ratio of the density of the material to that of a standard material, usually water. If a substance's relative density to water is less than one, it will float in it. Temperature and pressure have an impact on a substance's density. This variation is frequently not very noticeable for solids and liquids, but it is very noticeable for gases. As pressure is applied, an object's density rises, which reduces the object's volume. With a few rare exceptions, as temperature increases, a substance's density decreases as its volume grows.

To know more about density please refer: brainly.com/question/15164682

#SPJ4

3 0
1 year ago
An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the
elena-14-01-66 [18.8K]

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

F_c = F_g

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

Where

m = mass spacecraft

v = velocity

G = Gravitational Universal Constant

M = Mass of earth

r \rightarrow R+h \Rightarrow Radius of earth and orbit

Re-arrange to find the velocity

\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}

\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}

v^2_{orbit}=\frac{GM}{r}

v_{orbit} = \sqrt{\frac{GM}{r}}

v_{orbit} = \sqrt{\frac{GM}{R+h}}

Replacing with our values we have

v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}

v_{orbit} = 7676m/s

From the cinematic equations of motion we have to

t = \frac{d}{2v_{orbit}} \rightarrow Remember that the speed is double for the counter-direction of the trajectories.

Replacing

t = \frac{29000m}{7676m/s}

t = 3.778s

Therefore the time required is 3.778s

4 0
3 years ago
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