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scoundrel [369]

3 years ago

15

The statement that "The universe is the same everywhere and does not change over time" is known as _____.

2 answers:

kondor19780726 [428]3 years ago

8 0

The answer is B- Perfect Cosmological Principle

Sorry it took people so long to anser:)

djyliett [7]3 years ago

4 0

Answer: b. the perfect cosmological principle

Explanation:

The perfect cosmological principle is the extension of the cosmological principle. It states that the universe is homogeneous. It looks same from everywhere. It will look the same. It will never change over time instead of some of the gross changes occurring in the universe.

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At a certain altitude above the Earth's surface, the electric field has a magnitude of 135 V/m. How much energy is stored in 1.0

Paul [167]

Answer:

Explanation:

Given that,

Electric field E=135V/m

Energy stored in 1m³of air=?

The energy stored in an electric field is given as

u = ½ εo E²

Where

U is the energy stored

εo is permissivity and it value is 8.85×10^-12C²/N..m²

And E is the electric field

Then,

U=½×8.85×10^-12×135²

U=8.06×10^-8J/m³

Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³

4 0

3 years ago

Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea

lesantik [10]

Answer:

The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.

Question:

Why did you lie about being in college?

6 0

3 years ago

Read 2 more answers

A street light is on top of a 9 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 4 feet per second. At

Gekata [30.6K]

Answer:2 ft/s

Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

$\frac{x'}{x+x'}=\frac{3}{9}$

3x'=x+x'

x=2x'

and it is given $\frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s$

Differentiating

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} x'}{\mathrm{d} t}$

$4=2\times \frac{\mathrm{d} x'}{\mathrm{d} t}$

$\frac{\mathrm{d} x'}{\mathrm{d} t}=2 ft/s$

6 0

3 years ago

In the kitchen of your house you would like to run a 240 W blender, a 30 W phone charger, a 720 W toaster oven, and two 120 W li

meriva

Current pulled by these appliances is 10.25 A

Explanation:

Step 1: Find the total power used by these appliances.

Total power = 240 + 30 + 720 + (2 × 120) = 1230 W

Step 2: Given voltage is 120 V, find the current by using the formula Power = Voltage × Current

⇒ Current(I) = P/V = 1230/120 = 10.25 A

8 0

3 years ago

A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta

Vika [28.1K]

Answer:

$t_t=4.131\ s$

Explanation:

Given:

height above the horizontal form where the ball is hit, $y=1\ m$

angle of projectile above the horizontal, $\theta=30^{\circ}$

initial speed of the projectile, $u=40\ m.s^{-1}$

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

$u_y=u.\sin\theta$

$u_y=40\times \sin30^{\circ}$

$u_y=20\ m.s^{-1}$

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

$v_y=0\ m.s^{-1}$

Using eq. of motion:

$v_y^2=u_y^2-2g.h$ (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

$0^2=20^2-2\times 9.8\times h$

$h=20.4082\ m$ (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

$v_y=u_y-g.t$

$0=20-9.8t$

$t=2.041\ s$

<u>Time taken to descent the total height:</u>

we've total height, $h'=h+y$ $=20.4082+1$

$h'=u_y'.t'+\frac{1}{2} g.t'^2$

during the course of descend its initial vertical velocity is zero because it is at the top height, so $u_y'=0\ m.s^{-1}$

$21.4082=0+4.9t'^2$

$t'=2.09\ s$

<u>Now the total time taken by the ball to hit the ground:</u>

$t_t=t'+t$

$t_t=2.09+2.041$

$t_t=4.131\ s$

3 0

3 years ago