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3 years ago

Why is "Creation-Science" not considered true science?

2 answers:

8 0

My answer would be because it is based on the belief that God created the Earth and cannot be proven by science but requires someone to believe that it is true

qwelly [4]3 years ago

7 0

You may not believe this, but I do. The true answer is because there is so much evolution in the world people are starting to fall for it, evolution is not true and is completely ridiculous. Creation Science is true, God created the Heavens and the Earth (Genesis 1:1), Creation Science explains how it is true and why you should not fall for believing evolution, it explains Science in Biblical way that is true. I love reading about Creation Science and I am interested in becoming a Creationist! I am sorry if I offended anyone in any way, I don't intend to, I am just telling you the right answer.

You might be interested in

How does the kinetic energy of an object relate to its mass and velocity?

Setler [38]

The kinetic energy of an object is directly proportional to its mass, and to the square of its velocity.

5 0

3 years ago

Two rams run toward each other. One ram has a mass of 49 kg and runs west

olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram $m_1=49\ kg$

The velocity of this ram is $v_1=-7\ m/s$

Mass of the second ram $m_2=52\ kg$

The velocity of this ram $v_2=9\ m/s$

They combined after the collision

Conserving the momentum

$\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]$

Momentum after the collision will be

$\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}$

Therefore, option (d) is correct

4 0

2 years ago

Many of the whales in the ocean rely upon tiny marine organisms, called plankton, for food. If all of the plankton suddenly died

exis [7]

Answer:

I believe the answer to be B.

Explanation:

Without food, the whales would die.

8 0

2 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

2 years ago

The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1

ahrayia [7]

Answer:

The temperature is $T = 168.44 \ K$

Explanation:

From the question ewe are told that

The rate of heat transferred is $P = 13.1 \ W$

The surface area is $A = 1.55 \ m^2$

The emissivity of its surface is $e = 0.287$

Generally, the rate of heat transfer is mathematically represented as

$H = A e \sigma T^{4}$

=> $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where $\sigma$ is the Boltzmann constant with value $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value

$T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

$T = 168.44 \ K$

7 0

3 years ago