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2 years ago

If I were to transmit a radio wave in our three dimensional world could a fourth dimensional “being” be able to receive it?

1 answer:

5 0

Depends. Are you talking about a mathematical 4th dimension (in which there is infinite dimensions) or some sort of etheral dimension (in which there is no scientific evidence for)

If you mean the first then yes. But it depends how these beings exist. From our understanding we only can theorize shapes in 4-d and if we assume that there is only one universe these "beings" arleady exist and thus any message in 3-d would be sent to them like a shadow ("flat").
If they exist in a alternate "plane" then you would need some method to transverse this plan and if u did, then we would easily be able to communicate, but we would at first sound like a wild animal. They either would ignore us, not understand or perceive us, or they would attempt to send back a signal (essential they are ET's)

IF you mean the second then thats some mystic stuff and its pretty creepy (although a fun read for me :P)

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A constant force pushes on a brick, as shown below. A constant force pushes on three bricks as shown below. 师

JulijaS [17]

Answer:

5 UNDER 5

Explanation:

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2 years ago

Differentiate between capital g and small g

NNADVOKAT [17]

Answer:

Capital (G) is a universal gravitation law. and small (g) is acceleration of gravity of the each (9.8m/s^2)

Explanation:

According to the web

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2 years ago

A plane is traveling at 80 m/s. To prepare for landing, itslows down at a rate of 0.25 ms squared for 120 seconds. Calculate the

Tanya [424]

Answer:

The speed of the plane after it decelerates is 50 m/s

Explanation:

Motion with Constant Acceleration

When an object gains or losses velocity in time, it acquires acceleration. If this value is constant, we can calculate the final velocity (or speed in scalar terms) as:

$v_f=v_o+at$

Where vf is the final speed, vo is the initial speed, a is the constant acceleration, and t is the time the acceleration is acting.

The plane is originally traveling at vo=80 m/s and it slows down at a constant rate of $a=-0.25\ m/s^2$ during t=120 seconds. Note we have added the negative sign to the acceleration because the plane is slowing down, i.e., the acceleration is against the speed.

Thus, the final speed is:

$v_f=80-0.25*120$

$v_f=80-30=50$

$v_f = 50\ m/s$

The speed of the plane after it decelerates is 50 m/s

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2 years ago

An underground cannon launches a cannonball from ground level at a 35 degree angle. the cannonball is shot with an initial veloc

user100 [1]

Answer:

Time = 1.75[s]; Distance traveled = 21.5 [m]; Max height = 15 [m]

Explanation:

First, we have to break down the velocity vector into the X & y components.

$(v_{x})_{0} = 15 * cos( 35)= 12.28[m/s]\\(v_{y})_{0} = 15 * sin( 35)= 8.6[m/s]\\\\$

To find the time t that lasts the ball of cannon in the air we must use the following equation of kinematics, in this equation the value of y is equal to zero because it will be proposed that the ball lands at the same level that was fired.

$y=(v_{y} )_{0}-\frac{1}{2}*g*t^{2} \\where:\\g=9.81[m/s^2]\\t = time[s]\\y=0[m]$

$0=8.6*t-\frac{1}{2}*9.81*t^{2} \\4.905*t^{2}=8.6*t\\ t=1.75[s]$

In order to find the distance traveled horizontally from the cannonball, we must use the speed kinematics equation in the X coordinate.

$x = (v_{x})_{0} *t\\x=12.28*1.75\\x=21.5 [m]$

In order to find the last value, we must bear in mind that when the cannonball reaches the maximum height, the velocity in the component y is equal to zero, and we can find the value of and with the following kinematic equation

$y = (v_{y})_{0} *t+\frac{1}{2} *g*(t)^{2} \\y = 0*t+\frac{1}{2} *9.81*(1.75)^{2}\\ y=15 [m]$

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2 years ago

Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.