Ask question

Ask question

All categories

3 years ago

13

help i am confused about this riddle.IF NOTHING IS FASTER THAN LIGHT HOW DID THE DARK GET THERE FIRST

2 answers:

Digiron [165]3 years ago

8 0

Darkness is the absence of light.

vodka [1.7K]3 years ago

5 0

Answer:

because darkness is the absence of light, so that means darkness was there before the light came! :)

Explanation:

You might be interested in

A net force of 20 N acting on a wooden block produces an

Ymorist [56]

Answer:

From the second law of motion:

F = ma

we are given that the force applied on the block is 20N and the block accelerates at an acceleration of 4 m/s/s

So, F= 20N and a = 4 m/s/s

Replacing the variables in the equation:

20 = 4* m

m = 20 / 4

m = 5 kg

5 0

4 years ago

The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and

Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

$\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m$

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

$d=\sqrt{(x-x_o)^2+(y-y_o)^2}$ (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

$d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m$

hence, the displacement of the airplane is 3.45*10^4 m

6 0

3 years ago

Two identical loudspeakers that are 5.00 m apart and face toward each other are driven in phase by the same oscillator at a freq

Lunna [17]

Answer:0.0982 m

Explanation:

Given

distance between two loudspeaker is 5 m

frequency (f )=875 Hz

speed of sound (v)=344 m/s

Let $x_0$ be the smallest distance moved by observer then

Position of observer w.r.t to first speaker is

$x_1=\frac{L}{2}-x_0$

Position of observer w.r.t to second speaker is

$x_2=\frac{L}{2}+x_0$

$\Delta x=2x_0$

For Destructive interference

$\Delta x=\left ( m+\frac{1}{2}\right )\cdot lambda$

For minimum m=0

and $\lambda =\frac{v}{f}$

$2x_0=\frac{v}{2f}$

$x_0=\frac{v}{4f}=\frac{344}{4\times 875}=0.0983 m$

8 0

3 years ago

A clay pot at room temperature is placed in a kiln, and the pot\'s temperature doubles. how much more heat per second is the pot

xeze [42]

Answer:

16

Explanation:

If we treat the pot as a black body, then:

q = σ T⁴ A,

where q is the heat per second radiated,

σ is the Stefan-Boltzmann Constant,

T is the absolute temperature,

and A is the surface area.

If the absolute temperature doubles, then q increases by a factor of 2⁴ = 16.

7 0

3 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

Vladimir [108]

Answer

given,

Radius of sphere = 6.38 × 10⁶ m

time = 1 day = 86400 s

$\omega = \dfrac{2\pi}{T}$

$\omega = \dfrac{2\pi}{ 86400 }$

$\omega = 7.272 \times 10^{-5}\ rad/s$

a) at equator

$v = R_E \omega$

$v = 6.38 \times 10^6\times 7.272 \times 10^{-5}$

v = 464 m/s

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)$

$a = 0.03374 m/s^2$

b) at a latitude of 61.0 ° north of the equator.

$R = R_E cos \theta$

$R = 6.38 \times 10^6\times cos 61^0$

$R = 3.093 \times 10^6 m$

$v = R \omega$

$v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}$

$v = 225 m/s$

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(3.093 \times 10^6 )$

$a = 0.01635 m/s^2$

4 0

3 years ago